In analogy with the notion of a bilinear map, say that a bi-elementary map is a function $$f:\mathcal{A}\times\mathcal{B}\rightarrow\mathcal{C}$$ for some structures $\mathcal{A},\mathcal{B},\mathcal{C}$ such that:
For each $a\in\mathcal{A}$ the map $l_a^f:b\mapsto f(a,b)$ is an elementary embedding of $\mathcal{B}$ into $\mathcal{C}$.
For each $b\in\mathcal{B}$ the map $r_b^f:a\mapsto f(a,b)$ is an elementary embedding of $\mathcal{A}$ into $\mathcal{C}$.
In particular, say that a structure $\mathcal{S}$ is malleable iff for some structure $\mathcal{T}$ there is a bi-elementary $f:\mathcal{S}\times\mathcal{S}\rightarrow\mathcal{T}$ (if it helps to make things more concrete note that we may as well take $\mathcal{T}$ to be some sufficiently saturated model of $Th(\mathcal{S})$). Malleability is fairly rare. For example, if $\mathcal{S}$ is malleable then $\mathcal{S}$ has only one $1$-type: if $\mathcal{S}\models\varphi(s)\wedge\neg\varphi(t)$ then for any $f:\mathcal{S}^2\rightarrow\mathcal{S}$ either $\mathcal{S}\models \varphi(f(s,t))$ in which case $l_s^f$ is not elementary as witnessed by $t$ or $\mathcal{S}\models\neg\varphi(f(s,t))$ in which case $r_t^f$ is not elementary as witnessed by $s$. On the other hand, nontrivial malleable structures do exist: for example, $f:\mathbb{Q}^2\rightarrow\mathbb{Q}: (x,y)\mapsto x+y$ is a bi-elementary map when we equip $\mathbb{Q}$ with the ordering alone.
My question is:
Is there a snappy characterization of malleability?
EDIT: Note that in fact every group is bi-interpretable (using a parameter) with a malleable structure - namely, its torsor reduct. Given a group $(G; *,\cdot^{-1})$, let $t: G^3\rightarrow G: (a,b,c)\mapsto (a*b^{-1})*c$; the group operation $*$ is then a bi-elementary (indeed, "bi-isomorphic") map $(G;t)\times (G;t)\rightarrow (G;t)$. Since groups can be quite wild, this means that malleability in fact implies very little tameness. And the same argument works for ordered groups as well.