$I_m - AB$ invertible if and only if $I_n - BA$ invertible

For part (2):

Let $_{}−$ be invertible and let’s consider the following matrix expression $(_{}−)(_{}−)^{−1}$ which you can verify simplifies to $BA$.

If that is true, then what is $(_{}−)[(_{}−)^{−1}+I_{n}]$?

You can easily construct a similar argument if you first let $_{}−$ be invertible.

As hinted in Carmichael's comment, we have that the following are equivalent:

(1) $(I_n-BA)$ is invertible.

(2) $1$ is not an eigenvalue of $BA$.

(3) $1$ is not an eigenvalue of $AB$.

(4) $(I_m-AB)$ is invertible.

You've proven that $(2) \iff (3)$, so just note that

$$(I-M)v \iff Mv=Iv=v$$

for any matrix $M$.