Prove this refinement on the uniqueness of the identity element in a group I'm doing a problem that asks me to prove the following refinement on the statement of the identity of a group.

Problem Statement
Let $G$ be a group with identity element $e$ and let $e', g \in G$. If $e'g = g$ then $e' = e$.
I know this statement is a refinement because the condition for identity has been relaxed to only require that an element act like the identity when multiplied by one element from one side. Rather than requiring for all $g \in G$, $e$ returns the element $g$ when multiplied from both sides.

My attempt at a solution
Let $e, e', g \in G$. We call $e$ the identity element because it satisfies $eg = ge = g$,  $\forall g \in G$. If $e'g = g$, then $e'g = eg$ and by cancellation we see that $e' = e$ and thus $e'$ is the identity element in $G$.

Questions/Concerns
My main concern is that I don't think I've really utilized the new conditions about only needing to be multiplied by one element on one side. My proof feels... lazy for lack of a better word but I'm a bit confused as to where these new, relaxed conditions come into play. Any help here is greatly appreciated.
 A: Your proof is correct, so not much to say about that. Let's instead talk about the feeling of laziness.
The point is you can afford to be lazy because of 'the group being a group', that what separates a group from more general 'sets with multiplication', is doing all the heavy lifting!
Compare for instance to the following statement:

If for some 2-by-2 matrices $E$ and $G$ we have that $EG = G$ then $E$ equals the identity matrix

This statement if false. The trick to cooking up a counterexample is to take $G$ to be a matrix of which the the two columns are both scalar multiples of the same vector $v$ and to take $E$ to be any matrix for which $v$ is an eigenvector with eigenvalue 1.
Now while this statement about matrices is false, the statement in your question is true and that is because it demands $g$ to be taken from a group. If for instance we would repeat the statement about matrices, but would specify that we were talking solely about invertible matrices (which form a group) then the statement would be true. (And my counterexample would no longer be a counterexample as $G$'s of the form described are never invertible)
So in summary: if we want to tell apart the identity element in a group from the other elements we can use the 'lazy' definition involving only one other element $g$. If we want to tell apart the identity element in a larger set with multiplication then we need the 'full' definition of $eg = g$ for all $g$ because other elements $e'$ might satisfy $e'g = g$ for specific $g$.
A way to think about this in the context of more general (but still associative) multiplication is saying

the identity element is the only element $e'$ so that $e'g = g$ for some invertible $g$.

Then depending on on how hard it is to find an example of en invertible element you can think about this characterization as more or as less lazy than the characterization involving $e'g = g$ for all $g$. (Of course in a group finding an invertible element is extremely easy as every element is invertible.)
