How to handle a Displacement Operator which parametrically depends on a second coordinate?

We consider a bilinearly coupled two-oscillator Hamiltonian of the following form
$$
H
=
dfrac{p^2_s}{2m_s}
+
V(x_s)
+
dfrac{p^2_B}{2m_B}
+
dfrac{m_B}{2}omega^2_B
left(
x_B
–
dfrac{c_B}{m_komega^2_B}
x_s
right)^2.
$$

Can one diagonalize this Hamiltonian by employing the regular unitary bath displacement operator $D(x^0_B)$ for the bath mode (B) defined as
$$
D(x^0_B)
=
e^{-text{i},x^0_B(x_s),p_B/hbar},
hspace{0.3cm}
x^0_B(x_s)
=
-dfrac{c_B}{m_komega^2_B}
x_s,
$$

which, however, depends parametrically on $x_s$? I thought about this problem in two different ways, which seem to contradict each other. First, when I naively act with $D(x^0_B)$ on $H$, i.e.,
$$
D^dagger(x^0_B)
H
D(x^0_B),
$$

I get in trouble as the displacement operator seems not to commute with the kinetic energy operator of the S oscillator $[D(x^0_B(s)),p^2_s]neq 0$. This leads to interaction terms with respect to momentum. In contrast, my second reasoning was, that if I assume $D(x^0_B)$ acts only on the subspace of the bath oscillator, i.e., $D(x^0_B)=1_Sotimes D_B(x^0_B)$, then I think that the parametric dependence on $s$ should not bring any problems. Where is my fault here?