Proof of The Identity Matrix is a Two-Sided Multiplicative Identity

Write $A=(A_{ij})$ for $i\in\{1,\dots,m\}$, $j\in\{1,\dots,n\}$, and write $I_n=(\delta_{ij})$ as in Identity Matrix, where $\delta_{ij}=1$ if $i=j$ and $\delta_{ij}=0$ otherwise.

By Product of Real Matrices, for all $i\in\{1,\dots,m\}$ and $j\in\{1,\dots,n\}$,

$$(A\,I_n)_{ij}=\sum_{l=1}^{n}A_{il}\,\delta_{lj}.$$

Every term with $l\ne j$ vanishes, and the term with $l=j$ equals $A_{ij}$; hence $(A\,I_n)_{ij}=A_{ij}$ for all $i,j$, so $A\,I_n=A$.

Likewise,

$$(I_m\,A)_{ij}=\sum_{l=1}^{m}\delta_{il}\,A_{lj}=A_{ij},$$

so $I_m\,A=A$. The square case is the case $m=n$. $\blacksquare$