A one-sided inverse is two-sided in a group
Solution 1:
Uniqueness is not necessary. If $ab=e$, then $b=be=bab$. Multiply on the right by $b^{-1}$, the right inverse of $b$. Then
$$\begin{align} e&=bb^{-1}\\ &=(bab)b^{-1}\\ &=bae\\ &=ba. \end{align}$$
Solution 2:
You do no use uniqueness and you do not need it: $ab=e\to (ba)b=be=b$. Multiplying by $b^{-1}$ gives $ba=e$.