Just for fun, I will also show that there exists an uncountable family of pairwise non-embeddable elementarily equivalent finitely generated simple amenable groups ... but fail to provide a single explicit example of a pair of such groups. Let $\mathcal{G}$ be the Polish space of f.g. groups. Using recent work of Grigorchuk-Medynets, there exists a Borel reduction $\varphi: 2^{\mathbb{N}} \to \mathcal{G}$ from $E_{0}$ to $\cong$; say, $x \mapsto G_{x}$. Let $L$ be the language of group theory and let $\psi: \mathcal{G} \to \mathcal{P}(L)$ be the Borel map $G \mapsto Th(G)$. Then there exists a fixed complete theory and a comeagre $X \subseteq 2^{\mathbb{N}}$ such that $Th(G_{x}) = T$ for all $x \in X$. Consider the Borel subset $Z = \varphi(X) \subseteq \mathcal{G}$. Define a Borel coloring $\theta: [Z]^{2} \to 2$ by $\theta(G,H) = 0$ iff $G$, $H$ are incomparable with respect to embeddability. Then there exists a Cantor set $C \subseteq Z$ such that $\theta$ is constant on $[C]^{2}$. Since each f.g. group has only countably many f.g. subgroups, it follows easily that $C$ is an uncountable family of pairwise non-embeddable elementarily equivalent finitely generated simple amenable groups.
For each $n \geq 1$, let $F_{n}$ be the free group on $n$ generators and let $\mathbb{Q}^{n}$ be the direct product of $n$ copies of the additive group of the rationals. Since $\mathbb{Q}^{2} \equiv \mathbb{Q}$ and $F_{2} \equiv F_{3}$, it follows that $F_{2} \times \mathbb{Q}^{2} \equiv F_{3} \times \mathbb{Q}$. Clearly $F_{3} \times \mathbb{Q}$ embeds into $F_{2} \times \mathbb{Q}^{2}$ and it is easily seen that $F_{2} \times \mathbb{Q}^{2}$ does not embed into $F_{3} \times \mathbb{Q}$. On the other hand, using the fact proof that elementary subgroups of free groups are free factors, it follows that $F_{3} \times \mathbb{Q}$ cannot be elementarily embedded into $F_{2} \times \mathbb{Q}^{2}$.
For each $n \geq 1$, let $F_{n}$ be the free group on $n$ generators and let $\mathbb{Q}^{n}$ be the direct product of $n$ copies of the additive group of the rationals. Since $\mathbb{Q}^{2} \equiv \mathbb{Q}$ and $F_{2} \equiv F_{3}$, it follows that $F_{2} \times \mathbb{Q}^{2} \equiv F_{3} \times \mathbb{Q}$. Clearly $F_{3} \times \mathbb{Q}$ embeds into $F_{2} \times \mathbb{Q}^{2}$ and it is easily seen that $F_{2} \times \mathbb{Q}^{2}$ does not embed into $F_{3} \times \mathbb{Q}$. On the other hand, using the fact that elementary subgroups of free groups are free factors, it follows that $F_{3} \times \mathbb{Q}$ cannot be elementarily embedded into $F_{2} \times \mathbb{Q}^{2}$.