For the sake of simplicity I will describe the problem with a three group structure.
Suppose there are three groups $G_1$, $G_2$ and $G_3$. Suppose also there is a binary map $M:G_1\times G_2\to G_3$ so that
- For all $g_2$ in $G_2$, $M(\_,g_2):G_1 \to G_3$ is a homomorphism.
- For all $g_1$ in $G_1$, $M(g_1,\_) : G_2 \to G_3$ is a homomorphism.
- For all $g_1$ and $g_1'$, $M(g_1,\_)*M(g_1',\_)$ is the homomorphism $M(g_1*g_1',\_)$ and similarly for all $g_2,g_2'$.
I have encountered this structure and am looking for references and results for this structure.
The third condition is implied by the first two. You can call such a map bilinear, although that term is more commonly used when $G_1, G_2, G_3$ are all abelian.
Note that this condition is stronger than it looks in the nonabelian case. We have
$$M(ab, cd) = M(a, cd) M(b, cd) = M(a, c) M(a, d) M(b, c) M(b, d)$$
but we also have
$$M(ab, cd) = M(ab, c) M(ab, d) = M(a, c) M(b, c) M(a, d) M(b, d)$$
from which it follows that $M(a, d) M(b, c) = M(b, c) M(a, d)$ for all $a, b, c, d$. In other words, the image of $M$ is necessarily an abelian subgroup of $G_3$, and so $M$ factors through the abelianizations of $G_1$ and $G_2$. This shows that there is a "tensor product" $G_1 \otimes G_2$ such that bilinear maps correspond to homomorphisms $G_1 \otimes G_2 \to G_3$, but it's just the tensor product of the abelianizations of $G_1$ and $G_2$.