Here, I attempted an alternative proof for a restricted case (when the operation is commutative)
Let us define a homomorphism $ \phi:G \rightarrow G $ such that, $\phi (x) =x^3 $
The operation being commutative, the function is a homomorphism. Now, $ker ( \phi) =\{e\}$, i.e. the mapping is injective. Since $\phi$ is one to one and a mapping on the set itself (same cardinality) , it is surjective. Hence, we are done here.
Now, for the original question, can we follow the same route with some modification? I like the Bezout's Identity argument, but I want to do it in a bit different fashion.
Proposition Let $G$ be a finite group and $n$ a positive integer. Then the map $f: G \mapsto G$ defined by $f(g)=g^n$ is a bijection if and only if gcd$(|G|,n)=1$.
Proof (sketch) Bézout yields $1=k|G|+mn$, for some integers $k, m$. Then $g=g^{k|G|+mn}=g^{mn}$. Hence if $g^n=h^n$, then $g=g^{mn}=h^{mn}=h$. So $f$ is injective and since $G$ is finite it must be bijective. Conversely, assume gcd$(n,|G|)\neq 1$. Then we can find a prime $p$ with $p \mid n$ and $p \mid |G|$. By Cauchy's Theorem there is a non-trivial $g \in G$ with order$(g)=p$. Then $g^n=g^{p \cdot \frac{n}{p}}=1^\frac{n}{p}=1=1^n$. Since $f$ is injective this yields $g=1$, a contradiction.