The Baer-Suzuki theorem (Aschbacher, "Finite group theory", Theorem 39.6) says that if $X$ is a $p$-subgroup of a finite group $G$ then either $X\le O_p(G)$ or there exists $g\in G$ such that $\langle X,X^g\rangle$ is not a $p$-group. Taking $p=2$ shows that if $X=\{1,t\}$ has order two and its conjugates generates $G$, and the product of any two conjugates of $t$ generate a $2$-group then $G$ is a finite $2$-group. This only answers the question if $G$ is finite, so any counterexample would have to be infinite.