I don't sharply disagree with Dr Neumaier's answer; it is indeed the case that entanglement may only be discussed for Hilbert spaces that are tensor products.

However, if the two parts of the well are sufficiently distant, this is nearly the case of your situation, too. When one looks at it in this approximate way, the answer is that the electrons – assuming that you only occupied one spin state, for example both electrons are spin up – are **not entangled**.

Why?

The Hilbert space with two widely separately wells that can store electrons is approximately the tensor product
$$ {\mathcal H} = {\mathcal H}_\text{left well} \otimes {\mathcal H}_\text{right well}$$

The two individual product Hilbert spaces are not quite completely well-defined: one doesn't want to discuss quantum field theory on a "region of space" due to the problems with the boundary conditions. 

However, as long as these boundary conditions are not a problem, the Hilbert space does factorize in this way, and so does the state you wrote:
$$|\psi\rangle = |\text{1 electron}\rangle_\text{left well} \otimes |\text{1 electron}\rangle_\text{right well} $$
The system is simply composed of two independent systems – two wells in two different regions – that are not correlated or entangled at all. A non-entangled state is defined as one that can be written as a tensor product and that's exactly what we can do here.

We don't violate the Pauli exclusion principle here in any way because in this approximate two-region description of the system, the binary quantum number "rough position" (which is either "near left well" or "near right well") plays the same role as the spin or other quantum numbers. The two electrons have different eigenvalues of "rough position" which is why they can be in exactly the same state when it comes to energy, spin, and all other quantum numbers.

If several electrons are in vastly different regions of space, the Pauli exclusion principle becomes inconsequential, of course: the electrons are effectively distinguishable by their location. But to be able to look at the situation in this way, I had to organize the Hilbert space as a tensor product of pieces that correspond to individual regions. If we organize the Hilbert space according to "individual electrons that may a priori be anywhere", we can't really talk about the entanglement at all because the total Hilbert space of many electrons isn't a tensor product of the individual electrons' spaces: it's the antisymmetrization of it.

The simplest, strict definitions of entanglement don't apply to such antisymmetrized tensor spaces. There's still a natural convention that if we have antisymmetrized (or symmetrized) tensor product Hilbert spaces, we still consider the antisymmetrization (or symmetrization) of a tensor product state to be a non-entangled state. This includes your state. Such a definition will tend to produce similar verdicts as the procedure based on the quantum field (composed of various regions) that I described above.

At any rate, you won't find any helpful way to argue that (and why) these two electrons are entangled. The question is either ill-defined or they are not entangled.