Before of all, I do some necessary definitions.
Let $G$ be a graph.
For each $X\subseteq E(G)$, we denote the graph $G[X]=(V(G),X)$ (observe that $G[X]$ is a subgraph of $G$).
Moreover, define by the $\operatorname{part}(G)$ a partition of $V(G)$ such that vertices pair $\{u,v\}$ belongs to the same part of $\operatorname{part}(G)$ if and only if $u$ and $v$ belong to the same connected component of $G$. For example, in the figure below we have the graph $G$ with $V(G)=\{a,b,c,d,e\}$.
Therefore, $\operatorname{part}(G)=\{V_1,V_2\}$ such that $V_1=\{a,b,c\}$ and $V_2=\{d,e\}$.
That is, $\operatorname{part}(G)$ has the parts $V_1$ and $V_2$.
We say that $\operatorname{part}(G)=\operatorname{part}(H)$ if any vertices $u$ and $v$ are connected in $G$ if and only if $u$ and $v$ are connected in $H$.
Follows the illustration:
I wish to show the following:
Let $G$ be a graph and $G'=G/e$ (contraction of $e$ in $G$) with $e\in E(G)$. If $\operatorname{part}(G'[A])=\operatorname{part}(G'[B])$, $G'[A]$ is a forest and $G'[B]$ is a forest, then $\operatorname{part}(G[A])=\operatorname{part}(G[B])$.
I think that the conectivity from $G'$ to $G$ changes only at the ends of $e$. Since $G'[A]$ and $G'[B]$ are forests, then the ends of $e$ are not connected in $G[A]$ and $G[B]$.
Is this sufficient to conclude that $\operatorname{part}(G[A])=\operatorname{part}(G[B])$? Is my reasoning right?