No, I don't think this is valid. First you have to prove that f is a "good" function. The only way I can see to prove this is to prove that inverses are unique. But if you have this, the desired conclusion $(a^{-1})^{-1}=a$ follows immediately from $a^{-1}a=aa^{-1}=1$; i.e. $a$ satisfies the condition to be the inverse of $a^{-1}$. Secondly, I don't see how to prove f is 1-1 and onto without knowing $(a^{-1})^{-1}=a$.