The spin tensor is used to give a continuum macroscopic model of a material with lots of teeny-tiny spinning sources inside, but which does not have any large-scale gross material motion, so no ordinary visible mechanical angular momentum. You should think of a ferromagnet, where the electrons in the ferromagnet are aligned, so there is a net macroscopic amount of angular momentum around the z-axis hidden inside the system, and it's hidden because the ferromagnet as a whole isn't spinning at all, it's just staying put.

In the famous Einstein de-Haas experiment, reversing the magnetic field on such a ferromagnet leads the magnetic electron spins to flip, and the entire ferromagnet to twist slightly on a torsion pendulum so as to conserve the total mechanical plus spin angular momentum. This verifies both that the spinning electrons carry the same kind of angular momentum as rotating mechanical bodies, and also from the magnitude of the effect that it is the electron spin alignment that is responsible for the magnetism. This was the definitive experiment that established that electron spin is really spinning electrons, not just a two-quantum-state something or other.

So whenever you think of a macroscopic spin tensor, you should keep the Einstein deHaas ferromagnet in your head. If the aligned spin is in the z-direction, the angular momentum density in the stationary ferromagnet is all spin, there is no net macroscopic momentum density, and in some ordering of the indices, S^t_{xy} is the nonzero spin-tensor component of interest--- that's the density (t component) of the spin angular momentum in the x-y plane (xy components).

Einstein-Cartan shows how this macroscopic configuration gravitates. Inside the ferromagnet, macroscopically, there is torsion, but this torsion is zero everywhere outside the ferromagnet. You could today understand this completely intuitively by modelling the spin-density using a continuous fluid of infinitesimal Kerr black holes (see below). Considering Einstein's work on frame-dragging preceded this, I am not sure this isn't exactly how he thought of it, although the Kerr solution was unavailable then of course.

If you do this, you see that the Einstein Cartan theory is simply the consistent extension of GR so that a spinning continuum gravitates the same way as a continuum made of infinitesimal Kerr sources, and using Einstein-Cartan, you don't need to take the limit by hand, you can just formulate the equations with a spinning source. It is such a conservative extension of GR, that it's really what people mean by GR nowadays.

In short, eveything you said in the question is right. For the specific questions:

1. The spin tensor models materials with intrinsic spin, like a ferromagnet. The intrinsic spin is stored by spinning particle sources internally, and is not seen in the macroscopic mechanical stress-energy tensor, so that the total angular momentum includes a continuous density of contributions beyond r cross p (as in the Einstein de-Haas case, where p=0).

2. Yes, there is no spin tensor when there are no spinning particles (or fields with indices, vector, spinor, or tensor fields), the spin angular momentum is the difference of the total angular momentum $L^{\mu\nu\sigma}$ in the theory, and the mechanical angular momentum $x^\mu T^{\nu\sigma} - x^\nu T^{\mu\sigma}$, where T is the naive mechanical stress tensor. If there is no spinning matter, the difference is zero.

I should add that the case of spinning fields, i.e. fields with indices, is subtler, because with the appropriate field theory definition worked out by Belinfante, this tensor difference is zero again, the whole field angular momentum is again the naive r cross p angular momentum determined from the Poynting stresses. But this is best understood in quantum field theory, because there are actual spinning particles in the real world. All the Belinfante is showing is that if you choose the T right in the field theory, this angular momentum shows up as a Poynting style rotating field momentum for the case of continuous fields, although this doesn't work for describing spinning electrons by a classical field with a Poynting-like spinning field momentum because the Dirac equation (which can be modified for GR) doesn't have a classical field limit.

3. Yes, the two theories are identical when there are no spinning matter sources. They are also have identical vacuum equations outside spinning matter, where the torsion vanishes. I would go so far as to say that they are really the same theory, Einstein-Cartan is just showing what spinning matter does in standard GR. Einstein knew about ferromagnets.

The easiest way to get comfortable with the emergence of torsion in spinning solutions today, after Kerr, is to consider parallel transport of vectors around an infinitesimal square containing a Kerr black hole at the center of even more infinitesimal mass, and then adding up lots of spinning infinitesimal sources to a density. If the black hole is spinning along the z-axis, and you take an x-pointing vector along the y axis, so that the square surroundes the origin, where the black hole is, there is a clear failure of the symmetry relation for connection coefficients, i.e. the transport of the x-component in the y-direction is not equal to the transport of the y-component in the x-direction, and the failure is in the time direction.

To picture the torsion condition, consider a vector pointing in the x-direction being dragged by parallel transport an infinitesimal distance along a vector pointing along the y direction, so as to fill out a square as the dragging path. The classical interpretation of this dragging in the original Einstein-era variational-geometric point of view is that you have a bunch of parallel geodesics, starting at each separate point along the infinitesimal y-vector, which get no closer and no further away to leading order. The "No torsion" condition means geometrically that you fill out the same infinitesimal square (to leading order) by dragging the y-axis vector by parallel transport along the x axis vector, i.e. you fill out the same 2-d patch of points in spacetime no matter which way you drag the vector, x along y or y along x.

This relation fails in spacetimes with an infinitesimal spin at the center of the square because of the frame-dragging around the spin. The x-vector you are dragging in the y-direction becomes an infinitesimal geodesic of constant length. At the beginning of the transport, the x-geodesic is (say) going along the direction of the spin so that it is temporally helped along by the frame dragging in the spinning spacetime. After it crosses the Kerr-center, above the complicated central region, the geodesic in the same direction is going against the direction of the spin, so it ends up with time component deflected. By contrast, the relevant y-geodesic, the other side of the square, is crossing the spin in the opposite sense, so that at first it is going against the spin, and after crossing the Kerr-center it is going along with the spin, so that it gets a time-component deflected the other way. This means that the square fails to close into a proper 2-d sheet, the two transport paths are torn apart in the time direction, and this is the Einstein-Cartan torsion showing up as a global mismatch in the Kerr solution. The failure to close up a 2d sheet is clearly proportional to the spin-density, which in this case is a delta function at the origin with magnitude equal to the angular momentum L of the black hole. $S^t_{xy}$ is nonzero, and that's an asymmetric time parallel transport component when comparing transport of x and y vectors around a spinning center.

The general case can be understood from this by limits and superposition, just from the slowest falloff components of the Kerr metric. The difference is extremely intuitive, it is the difference in time between an observer that flies by against the rotation and one that flies by with the rotation, and it is due to the tilt of the time axis away from the spinning center just by the Machian frame-dragging of the spin.

To comment on the manipulations in the question, you should complete the solution of the EC equation of motion explicitly for the torsion. In d dimensions:

$$ {1\over \kappa} T_{ab}^c = \sigma_{ab}^c + {\delta^c_a \sigma_{br}^r - \delta^c_b \sigma_{ar}^r \over (2-d)} $$

The first term is what you get from the Kerr business above, it gives that the torsion is proportional to the spin-density. There are no traces for Kerr angular momentum, as the density is a t-component of x-y angular momentum. This generalizes by covariance to:

$$ {1\over \kappa} T_{ab}^c = \sigma_{ab}^c $$

Supplemented with the additional covariant condition that all the spin angular momentum is due to spinning particles $\sigma_{ab}^b = 0$, that it is Kerr-like.

This is the right physical condition for everything in the universe, I can't imagine the situation with a nonzero trace of $\sigma$, as this is not a superposition of spinning particles (which are traceless sigma). Moving to the rest frame of the density, where the top component is pure time, such a thing would be a time-space angular momentum, which by Noether's theorem is not an intrinsic angular momentum at all, but an intrinsic center of mass location. Unlike hidden intrinsic spin, there is no intrinsic hidden center of mass, but I suppose Einstein and Cartan gave the right equation to deal with such a situation, although how it would ever arise is beyond me, it wouldn't arise from spinning matter, that's for sure. I would just reduce the equation to the physical traceless case, as the case with nonzero trace for $\sigma$ looks purely mathematical, although if someone has an example where there is such a thing as "intrinsic center of mass location" I would be happy to hear it.