Identity is Unique

Theorem
Let $\left({S, \circ}\right)$ be an algebraic structure that has an identity element $e \in S$.

Then $e$ is unique.

Proof 1
Suppose $e_1$ and $e_2$ are both identities of $\left({S, \circ}\right)$.

Then by the definition of identity element:
 * $\forall s \in S: s \circ e_1 = s = e_2 \circ s$

Then:
 * $e_1 = e_2 \circ e_1 = e_2$

So $e_1 = e_2$ and there is only one identity after all.

Proof 2
Let $e_S$ be an identity of $\left({S, \circ}\right)$.

Then by definition, $e_S$ is both a left identity and a right identity.

By More than One Left Identity then No Right Identity, if there's more than one of either, there can't be one of the other.

So there can be only one of each, and by Left and Right Identity the Same, they are one and the same thing.