By Fermat's little theorem, we know if $p$ is a prime number, we have: $$a^p\equiv a\,\,\,(\text{mod }p).$$ We know in a finite field $F$ of order $p$ when $p$ is prime, characteristic of $F$ is $p$. So, using Fermat's little theorem $a^p=xp+a$ for some $x$. This means $a^p=a$ in $F$. This shows that all elements of $F$ satisfy the polynomial $X^p-X=0$. As this equation has $p$ roots, we conclude all the elements of $F$ are roots of this equation or conversely all the roots of this equation are elements of $F$.
At this point, I appreciate any corrections if I have any mistakes in the previous claims.
Now, a more general claim is that for every field $F$ of order $p^n$ when $p$ is prime, every element of $F$ satisfies the polynomial $X^{p^n}-X=0$. But I can't understand how this is concluded. Because, here $p^n$ is not a prime number. Any explanation of why this is correct is apprecaited!
Since you basically asked for a group theory free proof, here is one. (Of course, it's just group theory in disguise.)
Lemma. Let $F$ be a finite field with $q$ elements. Then $a^q=a$ holds for all $a \in F$.
Proof: For $a=0$ this is clear. So let $a \neq 0$. We consider the product of all the non-zero elements in $F$. $$s := \prod_{\large 0 \neq b \in F} b.$$ This means that when $F = \{b_1,b_2,b_2,\dotsc,b_q\}$ and $b_1 = 0$, we let $s = b_2 \cdot b_3 \cdots b_q$. Since $F$ has no zero divisors, $s \neq 0$. Now, the map $b \mapsto a \cdot b$ is a bijection from the set of non-zero elements of $F$ to itself. In fact, $b \mapsto a^{-1} \cdot b$ is inverse to this map. This means that we can make a substitution in the product. We get $$s = \prod_{\large 0 \neq b \in F} a \cdot b$$ There are $q-1$ factors, in each one we can take out $a$. So we get $$s = a^{q-1} \cdot \prod_{\large 0 \neq b \in F} b = a^{q-1} \cdot s$$ Since $s \neq 0$ and $F$ has no zero divisors, we get $1 = a^{q-1}$. Hence, $a = a^q$. $~~~\square$