Proof of $(ab)^{-1} = a^{-1}b^{-1}$ I wonder if you would consider this proof of my hypothesis of
$(ab)^{-1} = a^{-1}b^{-1}$ with $a,b\ne0$ correct.
By definition of the inverse of $ab$: $$(ab)^{-1}(ab)=1$$
Multiplying the equation with $a^{-1}$ and $b^{-1}$ gives $$(ab)^{-1}aba^{-1}b^{-1} = 1a^{-1}b^{-1}=a^{-1}b^{-1}$$ using property $1x=x$.
On the left hand side I rearrange the terms using the property $xy=yx$ to $$(ab)^{-1}aa^{-1}bb^{-1}= a^{-1}b^{-1}$$ which gives by using again the definition of the inverse $$(ab)^{-1}1\cdot1= a^{-1}b^{-1}$$
Using the property $1x=x$ on the left twice I have the answer.

Yes, this is correct. You could even do this without assuming commutativity, but then you would get
$$(ab)^{-1} = b^{-1}a^{-1}.$$
