I already know how to prove this using the definition of inverse and the associative property of matrix multiplication, but I was wondering if this would also be a valid proof.
As $A$ and$ B$ are $n \times n$ matrices then $AB$ is an $n\times n$ matrix.
Suppose $AB$ is invertible, then by the Fundamental Theorem of Invertible Matrices $AB$ is a product of elementary matrices. Then $A$ and $B$ must both be products of elementary matrices. By the same Fundamental Theorem this implies $A$ and $B$ are invertible.
Best Answer
The way to fill the gap in your proof is as follows: Let $AB=E_1E_2 \dotsb E_k$, where $E_i$ are elementary matrices. Then consider the homogeneous system $B\vec{x}=\vec{0}$. If this had a solution $\vec{x}_0$, then that will also be a solution to the homogeneous system $AB\vec{x}=\vec{0}$. But $AB$ is invertible so the only solution must be the trivial solution. This means $B$ is invertible. Now we can say that $B$ is a product of elementary matrices. Once we have that then we can write $A=(AB)B^{-1}$ to claim that $A$ is also invertible and product of elementary matrices.