Prove a nonempty set $G$ with an associative operation $\ast$ is a group iff the following equations are satisfied $y \ast g = h$ and $g \ast x = h$

Question

I'm currently working on an exercise and the body of the text for the exercise is as follows. I have a first draft of the proof but am missing some things and am unsure about some things as well so any criticism and guidance is welcome.

Prove a nonempty set $G$ with an associative operation $\ast$ is a group iff $\forall g,h \in G$, $\exists x,y \in G$ such that $$g \ast x = h$$ $$y \ast g = h$$ Hint: Prove that if $e \ast g =g$ for some $g \in G$ then $e \ast h = h$ for all $h \in G$. Then appeal to the problem that a nonempty set with associative operation and a existence of left identity and inverse implies it is a group.

Being that this is an if and only if statement we have to prove the forward and reverse statements. I will start by proving the reverse.

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Reverse Statement

Assume that $(G, \ast)$ is a group. Then for the equations $$y \ast g = h\ \text{and}\ g \ast x = h$$ we can take $y$ to be the left identity $e$ and $x$ to be the right identity $e$. So that we have $$g \ast e = g \in G\ \text{and}\ e \ast g = g \in G$$ Similarly we can take $x$ to be the right inverse of an arbitrary element $g$ and $y$ to be the left inverse of an arbitrary element $g$ - since groups are closed under inverses - to see that $$g \ast g^{-1} = e \in G\ \text{and}\ g^{-1} \ast g =e \in G$$ Finally. since groups are closed under the products determined by the operation $\ast$ if we take $x,y$ to be any other element in $G$ then we are guaranteed to have a solution in $G$ $$g \ast x = h \in G\ \text{and}\ y \ast g = h \in G$$ Thus $\forall g,h \in G$, $\exists x,y \in G\ \text{such that}\ g \ast x = h \in G\ \text{and}\ y \ast g = h \in G$ $\square$

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Forward Statement

For this part I am having some trouble getting started. Can we just assume that since $G$ is a nonempty set and we have the equation $y \ast g =h$ then it must contain a left identity element? Otherwise how would we follow the hint? Furthermore, I'm not sure how to go from the statement "if $e \ast g =g$ for some $g \in G$" to the statement "then $e \ast h = h$ for all $h \in G$". Would we just argue that if $e$ fulfills the role of the identity for one element then it must for all elements by definition and uniqueness of the identity?

From there I think I would say that the existence of the left identity element implies the existence of the left inverse for an arbitrary element since $e \in G$ and so by the equation $y \ast g = h$ we have $$g^{-1} \ast g = e$$ Next assume that $g \ast g = g$ then we have by multiplication of left inverse: $$g^{-1} \ast g \ast g = g^{-1} \ast g$$ Now by associativity of $\ast$ $$(g^{-1} \ast g) \ast g = g^{-1} \ast g$$ By definition of left inverses $$e \ast g = e \rightarrow g = e$$ Now, using the equation $g \ast x = h$ and the fact that $\ast$ is associative we can say that $$(g \ast g^{-1})(g \ast g^{-1}) = g \ast (g^{-1} \ast g) \ast g^{-1} = g \ast e \ast g^{-1} = g \ast g^{-1}$$ Now if we define $g \ast g^{-1} \equiv j$. Then we have, $j \ast j = j$ which implies that $j = g \ast g^{-1} = e = g^{-1} \ast g$. Thus right inverses are equivalent to left inverses. Now, we take $$g \ast e = g \ast (g^{-1} \ast g) = (g \ast g^{-1}) \ast g = e \ast g$$ That is, that right identities equal left identities. Then all we have left to prove is closure under products but the given equations show just that. $\square$