Prove that identity element is unique
As noted by MPW, the identity element $e \in G$ is defined such that $ae=a \quad \forall a\in G$.
While the inverse does exist in the group and multiplication by the inverse element gives us the identity element, it seems that there is more to explain in your statement, which assumes that the identity element is unique.
A more standard way to show this is suppose that $e, f$ are both the identity elements of a group $G$.
Then, $e=e \circ f$ since $f$ is the identity element.
$\quad\quad\quad= f\quad$ since $e$ is the identity element.
This shows that the identity element is indeed unique.
The definition of a group does not require that $a^{-1}\cdot a = e$ for every identity element $e$. It only requires that there be at least one identity for which that is the case.
It says:
- There is an identity element $e$ for which $e\cdot a = a\cdot e = a$.
- Every element $a$ has an inverse $a^{-1}$ for which $a^{-1}\cdot a = e$.
This does not rule out the possibility that there is another identity element $e_2$, which has the identity property $e_2\cdot a = a\cdot e_2 = a$, but for which $a^{-1}\cdot a \ne e_2$.
We could have $a^{-1}\cdot a = e \ne e_2$ and the axioms would still be satisfied.
For this reason, your proof is not correct.