The answer to your conjecture is yes, and you are completely right that the result for isometries implies that result for nonexpansive mappings (which I will simply all contractions here).
This follows from Sz.-Nagy's dilation theorem for contractions on Hilbert spaces, which says the following:
Theorem. Let $T$ be a linear contraction on a Hilbert space $H$. Then there exists a Hilbert space $V$ that contains $H$ and a unitary $U$ on $V$ such that \begin{align*} T^n = P U^nP \end{align*} for all integers $n \ge 0$, where $P$ denotes the orthogonal projection from $V$ onto $H$.
Now you can argue as follows:
Step 1. First assume that $\mathcal{H}_X = \mathcal{H}_Y$. Then $f$ is a contraction from a Hilbert space into itself, so we can apply the above dilation theorem. This theorem yields another Hilbert space $\mathcal{L}$ (which is the range of $\operatorname{id}-P$ in the theorem) and a unitary $U$ on $\mathcal{K} \oplus \mathcal{H}_X \oplus \mathcal{L}$ which is given by \begin{align*} U = \begin{pmatrix} A & B & F_1\\ C & D & F_2 \\ F_3 & F_4 & F_5 \end{pmatrix}. \end{align*}
Now you apply the known result for isometries to see that \begin{align*} \begin{pmatrix} D & F_2\\ F_4 & F_5 \end{pmatrix} + \begin{pmatrix} C \\ F_3 \end{pmatrix} \sum_{n=0}^\infty A^n \begin{pmatrix} B & F_1 \end{pmatrix} \end{align*} is an isometry. By projecting orthogonally onto $\mathcal{H}_X$ one concludes that $D + C \sum_{n=0}^\infty A^n B$ is contractive.
Step 2. Now let us consider spaces $\mathcal{H}_X$ and $\mathcal{H}_Y$ which might be different. Then define $\mathcal{H} = \mathcal{H}_X \oplus \mathcal{H}_Y$, as well as \begin{align*} \tilde B & = \begin{pmatrix} B & 0 \end{pmatrix} : \mathcal{H} \to \mathcal{K}, \\ \tilde C & = \begin{pmatrix} 0 \\ C \end{pmatrix} : \mathcal{K} \to \mathcal{H}, \\ \tilde D & = \begin{pmatrix} 0 & 0 \\ D & 0 \end{pmatrix} : \mathcal{H} \to \mathcal{H}. \end{align*} By applying what we have learned in Step 1 to the operator \begin{align*} \begin{pmatrix} A & \tilde B \\ \tilde C & \tilde D \end{pmatrix} \end{align*} we deduce that $\tilde D + \tilde C \sum_{n=0}^\infty A^n \tilde B$ is contractive. But this operator is equal to \begin{align*} \begin{pmatrix} 0 & 0 \\ D + C \sum_{n=0}^\infty A^n B & 0 \end{pmatrix}, \end{align*} so $D + C \sum_{n=0}^\infty A^n B$ is contractive, too.