If you interpret a $m×n$-matrix $A$ with entries in a ring $R$ as an $R$-linear map $f_A\colon R^n → R^m,~x ↦ Ax$, then $A$ is left-/right-invertible if and only if $f_A$ has the corresonding property.
As it turns out, for $f_A$ being right-invertible is equivalent to being surjective. For $f_A$ being left-invertible, there probably is a nice equivalence, too, but I can’t see it right now. You can at least say that $f_A$ needs to be injective and if you are working over a field, that’s sufficient as well!
Now, $f_A$ is surjective if and only if the columns of $A$ span $R^m$ and $f_A$ is injective if and only if the columns of $A$ are $R$-linearly independent.