Proposition. Suppose that $\forall a,\exists a^{-1}\textrm{ such that }a^{-1}a=1$, then $\forall a,aa^{-1}=1$, where $a^{-1}$ is the same as that in the hypothesis.

Proof. Let $a$ be arbitrary.
$$\begin{align*}a^{-1}a&=1\\ aa^{-1}a&=a\\ aa^{-1}aa^{-1}&=aa^{-1}\\ (aa^{-1})^{-1}aa^{-1}aa^{-1}&=(aa^{-1})^{-1}aa^{-1}\\ aa^{-1}&=1.\quad\square\end{align*}$$

It’s impossible to prove this proposition if we delete $\forall a$, because just $a^{-1}a=1$ is too weak and has nothing to do with the value of $aa^{-1}$.