This is not true, in general. For example, take $G = H = \mathbb Z$. Let $G \times H$ act on $\mathbb C^3$ in such a way that a generator of $G$ carries $e_1$ to $e_2$, and $e_2$ and $e_3$ to $0$, while a generator of $H$ carries $e_1$ to $e_3$, and $e_2$ and $e_3$ to $0$. It is easy to see that this representation is indecomposable, and that it is not of the form above.

This is not true, in general. For example, take $G = H = \mathbb Z$. Let $G \times H$ act on $\mathbb C^3$ in such a way that a generator of $G$ carries $e_1$ to $e_2$, and $e_2$ and $e_3$ to $0$, while a generator of $H$ carries $e_1$ to $e_3$, and $e_2$ and $e_3$ to $0$. It is easy to see that this representation is indecomposable, and that it is not of the form above.

[Edit:] as was, correctly, pointed out to me, my maps are not invertible; but you can make them invertible by adding the identity to both.

I had in mind the following. A representation of $G \times H$ corresponds to a $\mathbb C[x^{\pm 1}, y^{\pm 1}]$-module. The representation above corresponds to $\mathbb C[x,y]/((x-1)^2, (y-1)^2, (x-1)(y-1))$.