I haven't managed to solve it yet, but I've derived some principles that can help someone else solve it. (Note: I exclude 1 and any n>100 from the possible factor pairs and the term "number".)
As stratego said, X and Y (the hidden numbers) cannot both be prime. If they were, P would know them. S knows that P does not know them; the only reason this could be is that S knows they are not both prime.
Goldbach's conjecture states that any expression 2n, where n is an integer greater than 1, can also be expressed as j + k, where j and k are prime numbers. What this means is that for S to know for sure they are not both prime, the sum of X and Y cannot be any even number except 4 (in the case that X and Y both equal 2). Since the latter is plainly false (both professors would know if it were true), we can conclude that X + Y is an odd number.
Let us proceed. From S's knowledge of P's ignorance, P concludes that X + Y is odd. This enables P to identify the numbers. Therefore, XY has multiple pairs of factors, but only one of them has an odd sum.
What does this mean?
First, XY must be even, since the only way to have odd-summed pairs of factors is for X to be odd and Y to be even. An odd times an even produces an even. XY is therefore divisible by 2.
From this, we know that the even-summed pairs of factors contain only even factors, because two odd factors, while possessing an even sum, would have an odd product.
Therefore, of the possible pairs of factors of XY, only one contains an odd factor.
This might not be enough to determine the numbers yet, but let's look at S's last statement.
P's knowledge of the numbers tells S what the numbers are. This means that S is faced with several different pairs of numbers that sum to X+Y, but that only one of these pairs are uniquely odd-summed factors of their product. Hence:
S= A + B or C + D or E + F or ... BUT only the pair {A,B} are the only odd-summed factors of their product.
To sum up:
1. XY is even.
2. X + Y is odd.
3. X is the only odd number that, when multiplied by an even number, produces XY.
4. X and Y are the only pair that sum to X + Y for which the above is true.
That should narrow it a little, but I don't know exactly how to apply 3 and 4 as limiters. I can't seem to get any further right now, and I need sleep. Maybe TLC or another will pick it up tomorrow before I can get to it again.
