Let $G$ be a simple group and $H_1$ and $H_2$ be abelian subgroups of $G$ such that $H_1\cong H_2$ and $H_1,H_2\subseteq Cl_G(e_G)\cup Cl_G(x)$ for some $x\in G$, where $Cl_G(\cdot)$ denotes the conjugacy class on $G$.
Is it true that $H_1$ and $H_2$ are conjugate subgroups in $G?$ That is, is there is an element $g\in G$ so that $g^{-1}H_1g=H_2?$
Is it true that if $H_1$ and $H_2$ are finitely generated abelian subgroups?
As Derek indicated, there are counterexamples. Take $G=SL(2,16)$ and $x$ an involution. Then your union contains a full Sylow $2$-subgroup $P$, which is elementary abelian. Subgroups of $P$ are conjugate in $G$ if and only if they are conjugate in $N_G(P)$ by a theorem of Burnside. But there are too many subgroups of order $4$ to form a single conjugacy class (use the Gaussian binomial coefficient to compute their number).