That's an interesting question, but I can only think of an answer that renders the question trivial.

Before that, however, I would like to remark that you should not call it "self-energy". Your expression for $W$ (after integration over $\delta W$) is simply the energy contained in the electric field $\mathbf{E}$ generated by the charge density $\rho(x)$. The term "self-energy" specifically refers to the question of how the energy is to be interpreted for an isolated point particle, say an electron at position $x_0$, which corresponds to a charge density $\rho(x) = (-e) \delta(x-x_0)$. Does the electron "feel" its own electric field, thus contributing to the energy $W$? This seems impossible, because the integral is infinite in this case. Fortunately, in the case of *continuous* (or just *bounded*) charge distributions, the integral is well-defined and the "self-energy" problem is moot.

Now, concerning the question proper. There are two effects at play here: The electric field is generated ("emanates from") the charge distribution $\rho(x)$. This is Gauss' law. On the other hand, it also acts exerts a conservative force on the charge, i.e. the Lorentz law $\mathbf{F} = \rho\mathbf{E}$. These seem to be independent laws, changing either one of them will yield a different expression for the energy density. Your question is whether we can (at least formally) change these laws and obtain something that prohibits the definition of a total energy.

However, there is also the Lagrangian formulation of electrodynamics. Typically, a conservative system always has a Lagrangian formulation. (And this slight overgeneralization is what renders the question trivial, because you were only asking about the force being conservative, not the combined system force + charge, whatever that is exactly.) There, the interaction between electrons and field is mediated by an integral over a simple product $\int d^3x dt \, A_\mu j^\mu$. In particular, the contribution of the electric potential $\int d^3xdt\, A_0 \rho$. This term encodes *both* the force that the electric field puts on the charge *and* how the electric field is generated by the charge, as can be seen by functional differentiation with respect to $\rho(x)$ and $A_0$, respectively. So, regardless of how the other terms of the Lagrangian look like, this interaction energy will stay the same, and you will always get a well-defined energy for a charge density after solving for $A_0$.

Unfortunately, this answer seems circular, because the result we got out — the ability to define an energy density for the charge $\rho(x)$ — is something we already put in — the assumption that the physics are described by an interaction energy $\int d^3xdt\, A_0 \rho$ in the first place.