I think the experiment you are proposing is not possible in the way you want it.
Let us say we produce two photons in an electron-positron-annihilation with total momentum zero. (Since I don't see an easy way to produce entangled electrons I will talk about photons here, but I think it is not important for the argument). Those two photons are of course entangled in momentum: if one has momentum $\vec p$ the other one has momentum $-\vec p$.
But in order to make this statement you have to make a moemntum measurement on the initial state, i.e. know that the total momentum is zero with a certain $\Delta p$. But then, by means of the uncertainty relation, you only know the position where the photons were emitted with an uncertainty $\Delta x \propto (\Delta p)^{-1}$.
Now you can have two scenarios:
Either your double-slit is small enough and far enough away that due to the uncertainties $\Delta p$ and $\Delta x$ you do not know through which slit your photon goes. Or you still can tell (with some certainty).
In the second case there will never be an interference pattern. So no need for entanglement to destroy it.
But in the first case, due to the uncertainty $\Delta x$, measuring the position (by determining which slit your photon takes) does not give you an answer about the entangled photons position that is certain enough to tell which slit it will go through. Therefore you will see interference on both sides.
So an EPR like measurement is not possible in the experimental setup you propose.
I would assume that in general you need commuting observables, like spin and position in the Stern-Gerlach experiment, in order to measure EPR. But I didn't think that through yet.
addendum, 03-19-2014:
Forget about the second photon for a while. The first photon starts in a position state which is a Gaussian distribution around $\vec x_0$ and a momentum state which is a Gaussian around $\vec p_0$. After some time $t$ its position has evolved into a Gaussian of $\mu$ times the width around $\vec x_0 + \vec p_0 t$ (mass set equal to 1) while the momentum state is now $1/\mu$ times the width around $\vec p_0$. So while your spatial superposition gets larger - and thus better to measure with a double slit - the superposition in the momentum state, in which you have entanglement, gets smaller. You don't gain anything from entanglement, since your momentum wave-function is so narrow, that you know the momentum anyways.
It is actually not important to have space and momentum for this. Just take any non-commuting observables A and B, say with eigenstates A+, A-, B+, B-, and take two states S1 and S2 that are entangled in A. So measuring S1 in A+ implies S2 in A- and vice versa. But what you want is measure if S1 is in B+ or B- and from this conclude if S2 is in B+ or B-. And since A and B do not commute, measuring B with some certainty gives you a high uncertainty on A, meaning, for knowing if S1 is in B+ or B- you completely loose the information if it is in A+ or A-. So you cannot say anything about S2. On the other hand, as long as you are still in an eigenstate of A and know what to expect for the A measurement of S2, you don't know anything about the result of the B measurement.
So in order to do an EPR experiment you need entanglement in the observable you measure or an observable that commutes with it.
Please tell me if my thoughts are wrong.
