Let $x,y$ be integers greater than $1$, $P=xy$ and $S=x+y$.
Write $P=x_1\cdots x_n$, product of not necessarily distinct primes. If $n=2$, then necessarily $S=x_1+x_2$, so, if $S$ isn't the sum of two primes (this case), knowing $P$ tells nothing about $S$.
Then, we know that $n\ge 3$ and $S$ isn't the sum of two primes ($S$ isn't even, in particular, and then necessarily $P$ is even.).
So, necessarily $x$ is even and $y$ is odd. If we write $P=2^k p_1$, where $p_1$ is any prime, then necessarily $x=2^k$ and $y=p_1$, so, in this case $S$ would be known. Then, in this case, $P=2^k p_1\cdots p_m$, with $m>1$ and $p_i$ prime (since in this case knowing $P$ tells nothing about $S$).
So $S$ is odd and the set $\{xy:x+y=S,x,y>1\}$ contains one and only one number of the form $P=2^kp_1\cdots p_m$, with $p_i$ prime and $m>1$.
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In your link, the statement is a little different:
Sam (to Polly): "You can't know what x and y are."
Polly (to Sam): "That was true, but now I do."
Sam (to Polly): "Now I do too."
So, in this case we have $S$ odd, isn't the sum of two primes, and the set $\{xy:x+y=S,x,y>1\}$ contains one and only one number of the form $2^kp$, with $p$ odd prime, $k>1$, and $P=2^kp$, $x=2^k$, $y=p$. This excludes several possibilities and the rest is brute force (and we need an upper bound for $x$ and $y$).
Best Answer
There is in fact no solution to this puzzle. Martin Gardner realized this after presenting it in Scientific American:
For the puzzle to have a solution, there would have to be an admissible sum such that no matter how it is split up into two admissible addends, the resulting product also has at least one other admissible factorization. The following short Java program checks that there is no such admissible sum.
Note that the solution $(2,6)$ given on this page linked to by picakhu is incorrect. The sum is $8$, which could be the result of $(3,5)$, which has the unique factorization $3\cdot5$ that would allow B to deduce the sum $8$. The solution originally claimed by Martin Gardner was $(4,13)$.