I don't think this is equivalent. For if, for example, $G_A = G_B = \mathbb{Z}/2\mathbb{Z}$ (thought of as the set $\{0,1\}$), but we define $f(0) = 1$ and $f(1) = 0$, then I think $f$ is an isomorphism according to your definition, but it isn't actually an isomorphism.
First, it's not an isomorphism (according to the usual notion of isomorphism) because it doesn't map the identity to the identity, so it's not even a homomorphism.
On the other hand, $A = B = \begin{bmatrix} 0 & 1\\1&0 \end{bmatrix}.$
Then, one can check that $A_{00} = A_{11}$ and $B_{f(0)f(0)} = B_{f(1)f(1)}$, and similarly that $A_{01} = A_{10}$ and $B_{f(0)f(1)} = B_{f(1)f(0)}$.
Edit:
In fact, it's worse than I thought. In the example I gave, the 2 groups were actually isomorphic, but we have the following fact:
If $G_A$ and $G_B$ have the same number of elements, then they must be "isomorphic" according to your definition. Let $f:G_A\rightarrow G_B$ be any constant map. Then we have $B_{f(a)f(b)} = B_{f(x)f(y)}$ for any choices of $a,b,x,y$.
So, for example, $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}2\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z}$ are "isomorphic" according to your definition, but they aren't isomorphic at all.