Solution:

The answer is $19$. Indeed, suppose without loss of generality that $a > b$; then $a = b + p$ and $ab = b(b + p)$.

If now $p$ divided $b$ we could set $b = p b'$, and we would have that
$$ab = p b' (p b' + p) = p^2 b'(b' + 1)$$
would be a perfect square; but since $b'$ and $b' + 1$ are coprime, this would mean that both $b'$ and $b' + 1$ would be perfect squares, which is impossible (the only case in which this occurs among the integers, namely when $b' = 0$, is excluded by the hypothesis that $b$ is positive).

Therefore $a$ and $b$ must be coprime, and hence both perfect squares; we can therefore set $a = n^2$, $b = m^2$ with $n, m$ in turn positive. But from the fact that
$$p = a - b = n^2 - m^2 = (n - m)(n + m)$$
is a prime number it follows that one of $n - m$ and $n + m$ (and it will evidently be $n - m$) must be $1$. Hence $n = m + 1$, $a$ and $b$ are two consecutive squares and the sum $n + m$ of their square roots must be the prime number $p$.

However, since by hypothesis $ab \leq 10000$, $a \leq 100$ and hence $n \leq 10$, $m \leq 9$ and therefore $p = n + m \leq 19$; and precisely the case $a = 100$, $b = 81$, $n = 10$, $m = 9$, $p = 19$ turns out to fall within the hypotheses of the problem, hence $19$ is the maximum possible prime.