A binary operation * on a set is a mapping 𝑓:𝑆 × 𝑆 → 𝑆 that takes any two elements 𝑎,𝑏 ∈ 𝑆 to exactly one element 𝑓(𝑎,𝑏) ∈ 𝑆, which will be denoted 𝑎 ∗ 𝑏.
We know that by definition
For a set 𝑆 and a binary operation ∗ on 𝑆, if there is an element 𝑒 ∈ 𝑆 such that 𝑒 ∗ 𝑥 = 𝑥 ∗ 𝑒 = 𝑥 for all 𝑥 ∈ 𝑆, then we call 𝑒 an identity element of 𝑺.
Also know that any two elements of the set $S$ under the binary operation gives another element in the set $S$
since $e*x=x*e=x$ for any $x \in S$ and $e \in S$ then $e*e=e$ and therefore $e$ is its own inverse
Best Answer
Your proof is correct. "$b$ is an inverse of $a$" means that $a*b=b*a=e$. So the fact that $e*e=e$ tells us that $e$ is an inverse of $e$.
If the operation is associative, then it turns out that each element will have a unique inverse. Saying that $*$ is associative means that for all $a,b,c\in S$, $a*(b*c)=(a*b)*c$. Here is a helpful exercise:
Exercise Let $*$ be an associative binary operation on a set $S$ with identity $e$. Show that if an element $a\in S$ has an inverse, then the inverse is unique. In other words, show that if $b$ and $c$ are inverses of $a$, then $b=c$.
In your proof, you noted that $e*e=e$, which tells us that $e$ is an inverse of $e$. By the above exercise, we see that if the operation is associative, then $e$ is the inverse of $e$.
Solution to the exercise: Let $b$ and $c$ be inverses of $a$. Then