I hope that I am correctly interpreting the intended question.

I will use $x$ instead of $g$ to denote the distinguished element with the property that $[G,x] = G$, assuming there is such an element, and I will assume that $[G,x]$ denotes the subgroup $\langle [g,x]: g \in G \rangle,$ which is the usual group-theoretic interpretation of that notation.

Isn't it then the case that we have $G = [G,x]$ if and only if both $G$ is perfect, and $G$ is generated by the conjugates of $x$?

For if $G$ is generated by the conjugates of $x$, we have $G = \langle x \rangle [G,x],$ since $x^{g} = x(x^{-1}x^{g}) \in x[G,x]$ (and $[G,x]$ is a normal subgroup of $G$ by standard commutator identities).

Hence if $G$ is generated by conjugates of $x$ and $[G,x]$ is a proper normal subgroup of $G$, then $G$ has a non-trivial cyclic homomorphic image, and is not perfect.

Thus if $G$ is perfect, and is generated by the conjugates of $x$, then $G = [G,x].$

On the other hand, if $G = [G,x],$ then certainly $G = [G,G],$ so $G$ is perfect. Also, $G = \langle x^{-1}x^{g}: g \in G \rangle,$ so certainly $G = \langle x^{g} : g \in G \rangle$, and $G$ is generated by the conjugates of $x$.

In conclusion, a perfect group $G$ satisfies $G = [G,x]$ for some $x \in G$ if and only if $G$ is generated by one of its conjugacy classes.

notnormally generated by any single element. As already mentioned here, it is easy to check that no finite $G$ yields an example. $\endgroup$4more comments