$E(V,T)$ is an unspecified function. You don't need to know the form to get the derivation correct. The equation:
$$dE=\left(\frac{\delta E}{\delta V}\right)_T dV+\left(\frac{\delta E}{\delta T}\right)_V dT$$
is called the differential form of the total derivative and can be written for any function regardless of the form; it's simply a definition. As you noted in your edit, the subscripts on the derivatives indicate that they are to be taken with that variable held constant. So the first term on the RHS is the change in energy with change in volume for a constant temperature multiplied by the incremental change in volume.
If we know $dT = 0$ and we've established from the first law that $dE = 0$ then we are left with:
$$dE=\left(\frac{\delta E}{\delta V}\right)_T dV=0$$
Since the gas is expanding, we know that the incremental change in volume $dV \neq 0$. Since the product is zero (because $dE = 0$) then the only conclusion is that $\left(\frac{\delta E}{\delta V}\right)_T = 0$
WetSavanaAnimal aka Rod Vance has given a good introduction to the issues involved.
When I originally wrote an answer, I though that you were right: if you had a perfectly ideal system of perfectly hard spheres, with perfectly elastic collisions, in a container with perfectly rigid walls, and if all the particles started out with exactly the same speed, then the velocities could not evolve into a Maxwell-Boltzmann distribution, because I thought there was no process that could make the velocities become non-equal. However, I've realised I was wrong about that. For example, consider this collision:
The total $x$-momentum is zero but the total $y$-momentum is positive. This must be the case after the collision as well, so the motion must end up looking like this
with the top particle having gained some kinetic energy while the bottom particle loses some. Through this type of mechanism the initially equal velocities will rapidly become unequal, and the system will converge to a Maxwell-Boltzmann distribution just by transferring energy between particles, with no need for energy to enter or leave the system.
However, it's still possible to imagine special initial conditions where this won't happen. For example, we can imagine that all the particles are moving in the same direction, exactly perpendicular to the chamber walls, and are positioned such that they will never collide. A system in this configuration will remain in this special state forever, and not go into a Maxwell-Boltzmann distribution.
However, such a special state is unstable, in that if you start out with even one particle moving in a slightly different direction than all the others, it will eventually collide with another particle, creating two particles out of line, which can collide with others, and so on. Soon all particles will be affected and the system will converge to the Maxwell-Boltzmann distribution.
In reality, as Rod Vance said, in practice the walls will not be perfectly rigid and will be in thermal motion, which would prevent any such precise initial state from persisting for very long.
Even so, this seems to imply that the hard sphere gas system has at least one special initial state from which it will never reach a thermal state. But this isn't necessarily a problem for statistical mechanics. In this case (if my intuition is correct) the states with this special property form such a small proportion of the overall phase space that they can essentially be ignored, since the probability of the system being in such an initial state by chance is technically zero.
There can be cases where every initial state has a property like this. This means that the system will always remain in some restricted portion of the phase space and never explore the whole thing. But these are just the cases where there is some other conservation law, in addition to the energy, and we know how to deal with that in statistical mechanics.
People used to worry a lot about proving that systems were "ergodic", which essentially means that every possible initial state will explore every other state eventually. But nowadays a lot less emphasis is put on this. As Edwin Jaynes said, the way we do physics in practice is that we use statistical mechanics to make predictions and then test them experimentally. If those predictions are broken then that's good, because we've found new physics, often in the form of a conservation law. When the new law is taken into account, the new distribution will be seen to be a thermal one after all. So we don't need to prove that systems are ergodic in order to justify statistical mechanics, we just need to assume they are "ergodic enough", until Nature tells us differently.
Best Answer
At equilibrium, the temperatures must be the same, as you show in your first paragraph. This doesn't preclude temporary fluctuations while the system is out of equilibrium.
I agree with your reasoning that the gas within the blue region tends to heat up temporarily as work is done on it to push it into the empty side. (This aspect is discussed by Baker in "An easy to perform but often counterintuitive demonstration of gas expansion.") Correspondingly, the gas outside the blue region tends to cool down temporarily as it does that work. Over time, however, heat transfer causes the temperature throughout the container to equilibrate at the original temperature. Make sense?