First we observe that the statement holds if the set includes $a = 0$ or $b = 0$. Let us denote by $N(a, b) = a^5b^3 - a^3b^5$. Since $N(-a, -b) = N(a, b)$ and $N(-a, b) = N(a, -b) = -N(a, b)$, then WLOG we may assume that the 3 distinct integers are all positive. Now, it is easy to check that $a^5b^3 - a^3b^5$ is even and it is suffice to prove that $N(a, b)$ is a multiple of 5, which will certainly occur if either $a$ or $b$ is multiple of 5. Since
$$N(a, b) = a^3b^3(a^2 - b^2) = a^3b^3(a - b)(a + b),$$
what we have to prove is the following claim:
Given any 3 positive integers none of which is multiple of 5, the sum or difference of 2 of them is a multiple of 5.
Indeed, the last digit of any number not multiple of 5 lie in the set
$$\{1, 2, 3, 4, 6, 7, 8, 9\}$$
Let $A = \{1, 4, 6, 9\}$ and $B = \{2, 3, 7, 8\}$ (pigeonholes). Of the 3 integers (pigeons) in our set, by the PHP, at least 2 belong to $A$ or at least 2 belong to $B$, In any case, either their sum or their difference is a multiple of 5 as can be easily check, and we are done. $\square$