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 $\mathcal{K}$ 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.