Not quite done, but time has been scarce lately:
Proposition: In a finite group with non-cyclic abelianization, the subgroups generated by each conjugacy class are proper.
Proof: Suppose $G$ is a finite group whose abelianization $G/[G,G]$ is not cyclic. Let $x \in G$. Then $G/[G,G] \neq \langle x \rangle [G,G]/[G,G]$, so there is some maximal subgroup $M/[G,G]$ such that $\langle x \rangle [G,G]/[G,G] \leq M/[G,G]$. Since $G/[G,G]$ is abelian, $M/[G,G]$ is normal, and so $M \lhd G$ is normal, and so contains the subgroup generated by all conjugates of $x$. $\square$
Proposition: If $G/[G,G]$ is cyclic and $G$ is finite (or has a composition series), then there is a conjugacy class that generates $G$ as a subgroup.
Proof: This is a special case of theorem 4.7 of Berrick–Robinson (1993). $G$ itself has finite composition length, so Robinson's result says that $G$ is the subgroup generated by a single conjugacy class if and only if $G/[G,G]$ is. However, $G/[G,G]$ is generated by a single element (=conjugacy class for abelian groups) if and only if it is cyclic. $\square$
Proof: (Of conjecture, my attempt) Suppose $G/[G,G]$ is cyclic, generated by $x[G,G]$, and set $N$ to be the subgroup generated by the conjugates of $x$. Let $g \in G$. Then $x^g = x[x,g]$ so $[x,g] \in N$. Hence $x \in Z(G/N)$. ...
Counterexample: If $G/[G,G] = \langle x \rangle [G,G]/[G,G]$, then $G$ need not be the subgroup generated by the conjugacy class of $x$.
Proof: Take $G = \langle x \rangle \times Y$ with $Y$ non-identity perfect. Then the conjugacy class of $x$ is just $x$ itself, a proper subgroup. $\square$
Fix: the only problem is when $x$ is contained in a maximal normal subgroup not containing $[G,G]$, but in that case the quotient is non-abelian simple, and so a joe random element $y$ of that quotient should be such that $G$ is the subgroup generated by the conjugacy class of $xy$.