The answer is no. Let $G$ be a free abelian group of rank 2 generated by $x,y$. Let $S$ be the Meakin-Margolis expansion of $G$. It consists of all pairs $(X,g)$ with $X$ a finite connected subgraph of the Cayley graph of $G$ containing the origin and $g$. The product is $(X,g)(Y,h)=(X\cup gY,gh)$. The projection to $G$ is an idempotent pure homomorphism, so $S$ is E-unitary, and $S$ is generated by the edge from the origin to (1,0) and the edge from the origin to (0,1). Call these generators $x,y$ respectively. Now let $T$ be the quotient of $S$ by the relations $xx^{-1}=1=x^{-1}x$. It is not hard to see that $T$ is E-unitary since it is sandwiched between $S$ and $G$. Its elements can be viewed as pairs $(X,g)$ with $X$ a connected subgraph of the Cayley graph which contains the origin and $g$ with only finitely many vertical $y$ edges and containing the horizontal line through any vertex of $X$. Such graphs are precisely the Schutzenberger graphs of $T$.
These graphs in general do not quasi-isometrically embed as soon as they have a $y$ edge. For example of you take the Schutzenberger graph of $y$ you have the lines $x=0$ and $x=1$ and the edge from $(0,0)$ to $(0,1)$. So the distance from $(n,0)$ to $(n,1)$ in this graph is $2|n|+1$ while in the Cayley of $G$ the distance is $1$. So the embedding is not a quasi-isometry.