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.
The photon in transit between the light source and the screen is described by a wavefunction. Specifically, the wavefunction describes a photon that is delocalised i.e. it does not have a well defined position. Because the wavefunction is delocalised it encompasses both slits, which is why we say the photon goes through both slits.
Whenever you interact with this wavefunction you change it, and typically you will change it in such a way as to localise it. This is because the interaction happens at a point, and the result is to localise the photon at that point.
Now, when the photon interacts with the screen this does happen at a point, and indeed the interaction creates a spot on the photographic film/CCD/whatever at the point where the interaction happened. We can't say in advance where the interaction will occur, only that the probability of it happening is given by the wavefunction. So any one photon interacts at a point, but when we take many photons the points where they interact with the screen are distributed according to the wavefunction and together they build up the interference pattern.
However, if you interact with the photon before it has reached the slits then your interaction localises the wavefunction so the photon can no longer go through both slits. Because the wavefuntion immediately prior to the slits has changed, the effect the slits has on the wavefuntion changes as well and therefore so does the final pattern at the screen on the other side of the slits.
Best Answer
An interesting question, but by its own nature, the true answer is unknowable. If we managed to make some predictions about what happens to the unobserved slit, we would need to still experimentally verify them. But how could we? Any experiment that involves observing one slit and having zero knowledge about what happens at the other would necessarily require us to never observe the slit of interest. So we could never gain the data to verify our predictions. Unfortunately, when quantum mechanics says something is unknown, it usually means we can never know it or even guess at it.
To answer your last question, if you track when an electron is detected and the slit you are observing had none pass by, then deducing that it must have come through the other slit does count as observing it.