The solution to the restricted version still works, but it is no longer unique. I'll give a proof that the previous solution still works. According to Joe Z.'s answer, Summer is given $x+y=17$, and Proctor is given $x \cdot y=52$. Also, to avoid confusion I'm going to call Proctor male and Summer female.
Because $x \cdot y=52$, Proctor knows the numbers could be either 2 and 26 or 4 and 13, but he has no way to determine which they are.
Summer knows that since $x+y=17$, the numbers could be any of the pairs $(2,15), (3,14),(4,13),(5,12),(6,11),(7,10),$ and $(8,9)$. None of those pairs are two prime numbers, so she knew that Proctor wouldn't be able to tell what the numbers are based on their product.
Now that Proctor knows that Summer knew he wouldn't be able to tell the numbers by their product, he can reconsider the two possibilities - if the numbers we 2 and 26, then the sum would be 28. If the Summer had the number 28 then her list of possibilities would include the pair (5,23), both of which are prime. Then she would not have known that he would not be able to tell. So he then knows that the numbers are 4 and 13.
Now that Summer knows that Proctor knows the numbers, she can go through a similar process for her list of possible numbers.
- If the numbers were 2 and 15, then the product would be 30. Then Proctor's list of possibilities would be $(2,15), (3,10),$ and $(5,6)$. If this had been the case, then he would not have been able to eliminate $(5,6)$ because none of the pairs that sum to 11 are a pair of prime numbers. Thus Proctor would not have been able to know the numbers yet.
- If the numbers were 3 and 14, Proctor's list would include $(2, 21)$, which could not be eliminated.
- If the numbers were 5 and 12, Proctor would be unable to eliminate $(3, 20)$
- For 6 and 11, Proctor would be unable to eliminate $(2, 33)$
- For 7 and 10, Proctor would be unable to eliminate $(2, 35)$
- For 8 and 9, Proctor would be unable to eliminate $(3,24)$
By process of elimination, Summer knows that the numbers are 3 and 14.
As you can see from my analysis, at no point did the restriction of $x+y<100$ factor in to Summer or Proctor's ability to determine the numbers. This should actually be pretty obvious because for $x,y>1$, $x+y\le x\cdot y$, so $x\cdot y<100\rightarrow x+y<100$. In other words, Summer knows the sum, and Proctor knows the product is less than 100 so he also knows that the sum is less than 100, leaving the restriction as merely a help to people trying to solve the puzzle.
The only way the restriction would affect this would be if Proctor could eliminate one or more pairs because $N$ was less than the sum of one of the pairs he would not have been able to eliminate otherwise. For example, if $N=36$, then Proctor would be able to eliminate $(2, 35)$, which would mean 7 and 10 would have still been a valid possibility based on what Summer knew.
So I think that the question of generalizing this can be split into two parts:
- As $N$ gets larger, do more solutions become viable?
- For smaller $N$, do different solutions become viable?
EDIT: According to this page, there are at least three more possible solutions to this problem: $x=16,y=111$, $x=201,y=556$, and $x=421,y=576$. It's under the section "Sum-Product Puzzle: (Kiltinen/Young unbounded version)".