Yes, for $S_n$ the answer is obvious: None of the $S_n$ is decomposable hence there cannot be a reducible Coxeter system for $S_n$.

In general one can do the following construction: $W/W' = C_2^k$ where $k$ is the number of connected components in the graph that is obtained from the Coxeter graph of $(W,S)$ by deleting all edges $s-t$ with $ord(st)\in 2\mathbb{N}\cup\lbrace\infty\rbrace$.
In particular $W/W' = C_2$ for all finite Coxeter groups other than $B_n$ and $I_2(m)$ with even $m$. If there was a decomposition $W \cong W(X_1 \times X_2)$ for two nonempty Coxeter diagrams $X_1$ and $X_2$, then we would have $\dim_{\mathbb{F}_2} W/W' \geq 2$. This is a contradiction. Therefore we only need to care about the types $B_n$ and $I_2(m)$ with $m$ even.

If $m$ is even, then $I_2(m)$ is isomorphic to the Coxeter group of $A_1\times I_2(\frac{m}{2})$.

For $W(B_n) = (\mathbb{Z}/2)^n \rtimes S_n$ with uneven $n\geq 3$ the answer is also positive because the natural $\mathbb{F}_2[S_n]$-module $\mathbb{F}_2^n$ is decomposable: $C_2^n \rtimes S_n = X \times (Y \rtimes S_n)$ with $X=\langle(1,1,...)\rangle$ and $Y:=\lbrace x | \sum_i^n x_i = 0\rbrace$. This decomposition is a Coxeter group of type $A_1\times D_n$.

I think the answer should be negative for $W(B_n)$ with even $n\geq 4$ but I cannot think of a short argument at the moment. Here is what I got so far:

If $W(B_n) = N_1\times N_2$ then the projections of $N_i$ must be commuting, normal subgroups of $S_n$ that generate $S_n$ together. Ergo one of them projects onto 1 the other on $S_n$. Therefore we may assume wlog $N_1\leq (\mathbb{Z}/2)^n$ and for every $\sigma\in S_n$ there is a $(v_\sigma, \sigma)\in N_2$. Since $N_1$ is abelian, it acts trivially on itself. $N_2$ acts trivially on $N_1$ by assumption, hence $N_1\subseteq Z(W)=\langle(1,1,...)\rangle=X$.

The length-function of $B_n$ maps $(1,1,...)$ to zero since $n$ is even. The length-function is a non-zero homomorphism $N_1/N_1' \times N_2/N_2' = W/W' \to \mathbb{Z}/2$ and it maps $N_1$ to zero. Because we know $W/W' \cong (\mathbb{Z}/2)^2$ we get $\ker(l)/W' = N_1/N_1'$ and therefore $\ker(l) = N_1 \times N_2'$. Hence $v:=(1,1,0,...)$ lies in $N_1\times N_2'$. Therefore either $v$ itself or $v+(1,1,1,...) = (0,0,1,...)$ lies in $N_2$. The conjugates of both vectors generate $Y:=\lbrace x | \sum_i v_i = 0\rbrace$. Hence $N_1=X\leq Y\leq N_2$ and we finally have a contradiction: $W(B_n)$ is indecomposable if $n$ is even.

EDIT:
I totally forgot $F_4$. Well one can show that $W(D_4)$ is indecomposable as a group similar to the above proof: $W:=W(F_4) = W(D_4) \rtimes S_3$ where $S_3$ acts as the group of diagram automorphisms. In particular $s_1, s_2, s_3 s_2 s_3, s_4 s_3 s_2 s_3 s_4$ are a set of simple reflections for the $D_4$ subgroup and $s_3, s_4$ generate the $S_3$ subgroup.

If $N_1, N_2$ are direct factors of $W$, then wlog $N_2 \twoheadrightarrow S_3$ and $N_1\subseteq C_{N_2}(W) = \langle s_2, s_3 s_2 s_3, s_4 s_3 s_2 s_3 s_4 \rangle \cong C_2^3$ is abelian and hence central. But now we have $N_1\subseteq Z(W) \cap W(D_4) \subseteq Z(W(D_4)) = 1$.