Any Set with Associativity, Left Identity, Left Inverse is a Group

Your proof seems correct to me, and it also seems that you understood what is the problem with the axioms. The usual proof works just fine in the following case:

Let $(G, *)$ be a semi-group. Suppose
(1) $\exists e \in G$ such that $\forall a \in G,\ ae = a$;
(2) $\forall a \in G, \exists a^{-1} \in G$ such that for all $e\in G$ satisfying 1, $aa^{-1} = e$.

It is then obvious, in this case, that the element $e$ in (1) is unique: Indeed, since $G$ is nonempty (by (1)), let $g\in G$ be arbitrary. Then, if $e_1$ and $e_2$ satisfy (1), we have $e_1=gg^{-1}=e_2$.

The next case is more interesting:

Let $(G, *)$ be a semi-group. Suppose
(1) $\exists e \in G$ such that $\forall a \in G,\ ae = a$;
(2) $\forall e\in G$ satisfying (1) and $\forall a \in G, \exists a_e^{-1} \in G$ such that $aa_e^{-1} = e$.

The problem is to actually prove the uniqueness of the unit. Let's prove it:

Let $e$ and $f$ satisfy (1). Then $$f=ee_f^{-1}=(ee)e_f^{-1}=e(ee_f^{-1})=ef=e$$ Therefore, $e=f$, and we're actually in the first (and simpler) case.