Check my proof that $(ab)^{-1} = b^{-1} a^{-1}$

Your way is absolutely fine. As you note, there is in fact an easier way. It would be enough to show that the element $c$ such that $(ab)c = e$ is in fact $c = b^{-1} a^{-1}$:

$$\begin{align} (ab)b^{-1} a^{-1} &= a (b b^{-1}) a^{-1} \\ &= a e a^{-1} \\ &= a a^{-1} \\ &= e. \end{align}$$

These questions are standardly done by going straightforward, definition-based. So for the element $ab$ we seek the element $x$ s.t. $abx = xab = e$. Sure. We know such an element is unique (if not - prove this too).

So let's just do it. $ab b^{-1} a^{-1} = a (b b ^{-1}) a^{-1} = a e a^{-1} = a a^{-1} = e$. That's one direction.

$b^{-1}a^{-1} * ab = b^{-1} (a^{-1}a) b = b^{-1}b = e$

As for your method above - it looks great. Well done.

The definition of inverse is a*a-1 = I (ie a operated with inverse should give identity element) a-1*a = I (ie a inverse operated with a should also give identity element)

so here

$(AB) B^{-1} A^{-1} = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I$

$B^{-1} A^{-1} (AB) = B^{-1}(A^{-1}A)B = B^{-1} I B = B^{-1}B = I$

Here when $B^{-1}A^{-1}$ Operated to AB on both sides and in both case it given I (Identity Matrix ).

$X Z = I$ means that $Z$ is the inverse of $X$ Similarly If $ABB^{-1}A^{-1}$ is giving identity element then $B^{-1} A^{-1}$ is the inverse of $AB$.