The discrete spectrum is because the potential drops fast enough that you classically reach infinity in finite time. The system is bound around infinity, rather than bound at zero. The precise boundary conditions at infinity are defined by the analytic continuation, and make the spectrum well defined.

The finite energy motion has a velocity which is proportional to $\sqrt{E-V}$ ~ $x^2$, and the time it takes to classically go from 0 to infinity is $\int {1\over x^2} dx$ which is convergent. The integral for J is divergent:

$$ J = 2\int_{-\infty}^\infty p dx =2 \int \sqrt{2E+x^4} dx $$

This is infinite, which means that doing the phase-matching to define the bound states for finite energy is ambiguous. This ambiguity is resolved by finding the ground state, setting its energy to zero, and talking about J relative to the ground state. The derivative of J with respect to E, the inverse level spacing, is the classical period, and is finite.

$$ {\partial J \over \partial E} = 2\int {1\over \sqrt{2E+x^4}} dx = T_\mathrm{cl} $$

The Bender prescription picks you exactly which bound wavefunctions to use for this problem, by linking it with the Harmonic oscillator by this particular continuation. The prescription near $V=-x^4$ is $V = -x^4 + x^2(1 + i\mathrm{sgn}(x) \epsilon log |x|)$, which is a particular kind of circularizing boundary condition, where wavefunction is removed at one end (positive imaginary potential) and introduced at the other end of space (negative imaginary potential). Although this is locally non-unitary, you can make it unitary by reinterpreting the Hilbert space inner product.

The same sort of thing happens when you start Bender's deformation at V= x^2(ix)^{i\epsilon), you get a negative and postive imaginary contribution at the two ends of space, and the space of allowed wavefunctions in the Hilbert space deforms. But the imaginary part on the two ends is opposite in sign near $x^2$ and near $x^4$, the two ends of the circuit, so that the situation has smoothly moved the bulk of the Hilbert space wavefunction mass to lie at infinity rather than at zero, in this particular way.

If you ignore Bender, the standard answer for this potential is not really unbound, it is ambiguous, depending on boundary conditions at infinity. If you take a symmetric interval [-L,L] with L large, impose the potential plus a constant which normalizes the ground state energy to be zero, rather than negative infinity. With circular boundary conditions, then you get some silly spectrum which is *not* Bender's spectrum, and which depends on L strongly.

Benders spectrum comes from this particular analytic continuation, which defines its own Hilbert space. The space of allowed wavefunctions and operators, while still representable as functions on R, are not the usual square-integrable functions, but are a certain collection which are defined relative to the naive non-normalizable ground state of the potential. But the reason these potentials give a discrete spectrum is simply that you oscillate around and come back classically in finite time, the Bender prescription is simply one path which defines precisely how to link up negative infinity and positive infinity. There are other prescriptions which give different answers, as usual for analytic continuation.