Question: I tried getting the answers in similar questions, everyone says that it's not necessary, but if $e$ is the identity element for any binary operation $\*$, which is not associative and commutative, how can

$$a\*e=a=e\*a$$

when it is not commutative, i.e. $a\*b \ne b\*a$?

Even if we get a value by solving $a\*e=a$. Will we get the same value by solving
$e\*a=a$ ?
Please provide an example.

Answer: 
Actually, given *any* set $S$ and operation $\*$ on it (so possibly neither associative nor commutative), we can simply *extend* this with a new symbol $\color{red}0$ (i.e., $\color{red}0\notin S$) and on the set $S':=S\cup\{\color{red}0\}$ define an operation $\color{red}\*$ by
$$x\color{red}\*y:=\begin{cases}x&\text{if }y=\color{red}0\\
y&\text{if }x=\color{red}0\\x\*y&\text{otherwise} \end{cases}$$
Then $\color{red}\*$ is not associative/commutative if $\*$ is not associative/commutative. But $\color{red}0$ is neutral.
Isn't subtraction a binary function which has an identity ($x-0=x$) although it is not commutative ($5-0 > 0-5$) or associative ($5-(4-3) > (5-4)-3$)?