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 (the "big" Hilbert space doesn't constrain the fields near the boundaries around the wells at all while the smaller Hilbert spaces have to impose some boundary conditions, so the factorization above can't be exact).
However, as long as these boundary conditions are not a problem (for example because it's guaranteed that everything is almost totally confined near the well and nothing gets close enough to these boundaries), 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.
This extra quantum number is also the reason why you have two nearby energy low-lying states of the two-well problem. There's a two-dimensional Hilbert space for a single electron spanned by energy eigenstates with energies $E_1,E_2$: the corresponding eigenvectors are "even" or "odd" functions of the position (the wave functions either have the same sign in both wells or the opposite sign). In the approximation in which the space between the wells is impenetrable and the boundary conditions for the regions don't pose a problem, we have $E_1=E_2$ and the two-dimensional Hilbert space may also be generated from another basis containing the ground state of the left well and the ground state of the right well. In this approximation, we're just filling two states that only differ by the "rough position" by the maximum number of two electrons.
The inequality $E_1\neq E_2$ in your exact treatment only arises because there's a nonzero probability amplitude for an electron to tunnel from one well to the other one. If it couldn't tunnel, we would have the exact "doubling" of the Hilbert space for a single electron. For the same reason, one can't measure the energy "in one well only" with the accuracy needed to distinguish $E_1$ and $E_2$.
If your measurement apparatus is confined to the vicinity of one well, the error in your energy measurement can't be smaller than $E_1-E_2$ so you won't be able to say "which of the two nearby states" the electron is in. The same holds for the vicinity of the other well which is why the measurement in one well can't influence anything detectable near the other well.
The impossibility to distinguish $E_1$ and $E_2$ by a measurement near a single well is easy to prove; if you measure the electron near the left well, with whatever low-lying energy near $E_1$ or $E_2$, you are proving that this electron is in an eigenstate of the "rough position". But the operator of "rough position" doesn't commute with the total energy; the eigenstate $|\text{left well ground state}\rangle$ is a linear superposition of the $|E_1\rangle$ and $|E_2\rangle$ eigenstates (it's the right linear superposition that vanishes near the other well), something like $$ |\text{left well ground state}\rangle = \frac{1}{\sqrt{2}} \left( |E_1\rangle - |E_2\rangle \right ) $$ If you've measured the "rough position", you are totally uncertain about the eigenvalue of the "exact energy" because these two operators don't commute with one another; a textbook case of the uncertainty principle. If the two wells are equally deep etc., by seeing an electron near the left well, you have 50% odds that its energy was $E_1$ and 50% odds that it was $E_2$ and nothing can be changed about these odds because they follow from the displayed equation above.
In terms of operators, we may say that in the basis "left well ground state" and "right well ground state", the operator of "exact energy" looks like $$ H = \frac{E_1+E_2}{2}\cdot{\bf 1} + \frac{E_1-E_2}{2} \cdot \sigma_1 $$ where the second term is proportional to an off-diagonal matrix similar to the first two Pauli matrices. It isn't diagonal in this basis so if we know that we found an electron near the left well, we know that its "exact energy" (whether it is $E_1$ or $E_2$) is maximally uncertain. And vice versa. If we find an electron in the state $E_1$, and we are sure it is not $E_2$, then this electron must be in a wave function that is nonzero near both well, so we don't learn anything about the "rough position" (left or right) which remains maximally uncertain.
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. So the dimension of the Hilbert space for the two separated wells is the simple product of the dimensions of the Hilbert spaces for the individual wells; there's no additional "antisymmetrization" we should do here because we're discussing "off-diagonal blocks" of a matrix and the antisymmetric part of the state is hiding in the convention how we label the two electrons.
But to be able to look at the situation in this factorized 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. And even if you found a (contrived) definition that would allow you to say that the simple state is entangled, such an "entanglement" will have no physical consequences. Two highly separated regions (or wells) are independent. In particular, the laws of quantum field theory are exactly local so a measurement or decision done near one well won't immediately influence a spatially separated other well.