I am assuming two domains below: integers ($\mathbb{Z}$), & rationals($\mathbb{Q}$); with common identity element $1$.
The binary exponentiation operator yields for the simplest case of two integers, $a,b$: $c = a^b= a*a*\cdots a$ (b times).
An identity element means that on multiplying $a^b$ by something like $c$ as inverse yields $1$ element of the set.
So, $c = \frac1{a^b}\implies a^b = \frac1c$
$\implies$
Taking natural log on both sides $b = \frac1{ln(a)}\cdot ln(\frac1{c})$, so it is clear that integers are not capable to have inverse w.r.t. the exponentiation operator; only rationals can.
However, integers are closed w.r.t. the exponentiation operator.
Also, need help to convert the above argument for proving that inverse for the domain of rationals would exist, for the exponentiation operator.
Suppose you are dealing with the exponentiation operation.
I claim that there is no identity element, $e$ for this operation for the set $X$, where $X$ can be either $\mathbb{R}, \mathbb{Q}$, or $\mathbb{Z}$.
Suppose there is such element then we must have $\forall a \in X, e^a =a^e=a$. If such element exists, we know that it is unique. From, $a^e=a$, we know that $e=1$. However, this means that $1^a=a$, we reach a contradiction if $a=2$. Hence, the identity doesn't exists and there is no inverse to talk about.