My question was inspired by the recent post: The centralizer of an element x in free group is cyclic.
Is it true that non-identity elements have abelian centralizer in free metabelian groups?
This is probably very simple, but I couldn't find myself a proof.
I don't know if it helps but I was able to reduce the problem to the following assertion: $$\forall a\in F, \forall b\in F, [a\in (F-F'), b\in (F'-F'') \Longrightarrow [a,b]\notin F'']$$ (here $F$ is a free group (not metabelian)).
$\hskip250pt$ Thanks for any ideas!
Free metabelian groups have faithful embeddings into the affine group $K\rtimes K^*$ of some field $K$ (Magnus embedding), or equivalently matrices $\begin{pmatrix}a & b\\ 0 & 1\end{pmatrix}$ with $a\in K^*$ and $b\in K$, and the latter group has the property (straightforward exercise) that the centralizer of any nontrivial element is abelian. This passes to subgroups and answers your question.