Since I did not find any reference, I included the results in my recent preprint (in greater generality): Rotation groups virtually embed into right-angled rotation groups. I record below the sketch of the argument in the specific case of Coxeter groups.
Theorem. Let $\Gamma$ be a finite labelled graph and let $C(\Gamma)$ denote the corresponding Coxeter group. For all $\Phi,\Psi \subset V(\Gamma)$, the double coset $\langle \Phi \rangle \langle \Psi \rangle$ is separable in $C(\Gamma)$.
Lemma. For every $\Phi \subset V(\Gamma)$, $\langle \Phi \rangle$ is separable in $C(\Gamma)$.
Proof. In the Cayley graph of $C(\Gamma)$, $\langle \Phi \rangle$ is convex. Let $\mathcal{J}$ denote the collection of the walls tangent to $\langle \Phi \rangle$, i.e. disjoint from $\langle \Phi \rangle$ but not separated from $\langle \Phi \rangle$ by another wall. Let $\mathrm{Rot}$ denote the subgroup of $C(\Gamma)$ generated by the reflections along the walls in $\mathcal{J}$. It acts on $C(\Gamma)$ with $\langle \Phi \rangle$ as a fundamental domain. Then $\langle \mathrm{Rot}, \Phi \rangle$ yields a finite-index subgroup that splits as $\mathrm{Rot} \rtimes \langle \Phi \rangle$. Thus, $\langle \Phi \rangle$ is a virtual retract, and a fortiori separable since $C(\Gamma)$ is residually finite. $\square$
Lemma. For all walls $A$ and $B$, $\mathrm{Cross}(A,B):= \{ g \in C(\Gamma) \mid gA \text{ transverse to } B \}$ is separable in $C(\Gamma)$.
Proof. Let $a$ (resp. $b$) denote the reflection along $A$ (resp. $B$). Then, given an element $g \in C(\Gamma)$, $gA$ and $B$ are transverse iff $gag^{-1}$ and $b$ generate a finite dihedral group; otherwise, they span an infinite dihedral group. Therefore, fixing a number $N \geq 1$ divisible by all the labels from our graph $\Gamma$, we can write $\mathrm{Cross}(A,B)= \phi^{-1}(1)$ where
$$\phi : \left\{ \begin{array}{ccc} C(\Gamma) & \to & C(\Gamma) \\ g & \mapsto & (gag^{-1}b)^N \end{array} \right..$$
Since $\phi$ is continuous with respect to the profinite topology and that $C(\Gamma)$ is residually finite, we conclude that $\mathrm{Cross}(A,B)$ is separable in $C(\Gamma)$. $\square$
Proof of the theorem. If $\Phi = \Psi$, then the desired conclusion follows from our first lemma; so we assume that $\Phi \neq \Psi$. Construct a new labelled graph $\Gamma^+$ from $\Gamma$ by adding two new vertices $u_\Phi$ and $u_\Psi$; by connecting $u_\Phi$ (resp. $u_\Psi$) to all the vertices in $\Phi$ (resp. $\Psi$) with edges labelled by $2$; and by connecting $u_\Phi$ and $u_\Psi$ with an edge labelled by $2$. Let $J_\Phi$ (resp. $J_\Psi$) denote the wall in $C(\Gamma^+)$ crossing the edge $[1,u_\Phi]$ (resp. $[1,u_\Psi]$). It turns out that
$$\mathrm{Cross}(J_\Psi,J_\Phi)= \langle \Phi \rangle \langle \Psi \rangle \langle u_\Phi, u_\Psi \rangle.$$
Since $\mathrm{Cross}(J_\Psi,J_\Phi)$ is separable in $C(\Gamma^+)$ and since $C(\Gamma)$ is separable in $C(\Gamma^+)$ (once identified with the subgroup $\langle V(\Gamma) \rangle$), we conclude that
$$\mathrm{Cross}(J_\Psi,J_\Phi) \cap C(\Gamma) = \langle \Phi \rangle \langle \Psi \rangle$$
is separable in $C(\Gamma)$. $\square$