If $m≠n$ and we have the matrices $A$ $(m\times{n})$, $B$ $(n\times{m})$ and $C$ $(n\times{m})$ such that $AB=I(m\times{m})$ and $CA=I(n\times{n})$, does $B=C$?
I know the proof that it is true if we are talking about square matrices, but it doesn't help in this case.
The conditions on your matrices cannot be satisfied. If $m$ and $n$ are positive integers such that $m\neq n$, then there does not exist an $m$-by-$n$ matrix $A$ (over any base field $\mathbb{K}$), along with two $n$-by-$m$ matrices $B$ and $C$, such that $AB=I_{m\times m}$ and $CA=I_{n\times n}$. Here, $I_{k\times k}$ is the $k$-by-$k$ identity matrix.
To show this, we may assume without loss of generality that $m<n$ (otherwise, replace $A$ by $A^\top$, $B$ by $C^\top$, and $C$ by $B^\top$). Suppose on the contrary that such $A$, $B$, and $C$ exist. Since $A$ is an $m$-by-$n$ matrix, the rank of $A$ is at most $m$. Therefore, as the column space of $CA$ is a subset of the column space of $A$, the rank of $CA$ is at most $m$. This contradicts the assumption that $CA=I_{n\times n}$, as $I_{n\times n}$ is of rank $n>m$.
