To be totally clear: no, the decomposition as a representation of $A$ and the decomposition as a representation of $B$ separately don't determine the decomposition as a representation of $A \times B$, because this is not enough information by itself to determine which irreducibles of $A$ pair with which irreducibles of $B$ in general. 

The smallest counterexample is $A = B = C_2$ acting on a $2$-dimensional vector space $V$ such that, as a representation of either $A$ or $B$, $V$ decomposes as a direct sum of the trivial representation $1$ and the sign representation $-1$. This means that $V$ could be either $1 \otimes 1 + (-1) \otimes (-1)$ or $1 \otimes (-1) + (-1) \otimes 1$ (the $+$ here is a direct sum but I find writing direct sums and tensor products together annoying to read) and you can't tell which. You can construct a similar counterexample out of any pair of groups $A, B$ which both have non-isomorphic irreducibles of the same dimension.