You have already got correct answers proving it with abstract mathematics.
But I would like to complement these with a concrete example,
to show that the reasoning is actually very simple.
Let us take for example a single particle with spin $\frac 12$.
This can be described by the tensor product of a simple wavefunction
$\psi(x,y,z)$ with a $2$-component spinor $\begin{pmatrix}\psi_+\\\psi_-\end{pmatrix}$.
This means you have a $2$-component spinor wavefunction.
$$\psi = \begin{pmatrix} \psi_+(x,y,z)\\\psi_-(x,y,z)\end{pmatrix}$$
Then the $x$-operator is the multiplication by $x$
which obviously acts different for different positions $(x,y,z)$,
but in the same way on $\psi_+$ and $\psi_+$.
And the $S_y$-operator is a matrix multiplication operator
$$S_y=\frac 12 \hbar \begin{pmatrix}0&-i\\i&0\end{pmatrix}$$
which scrambles $\psi_+$ and $\psi_-$,
but does so independent of $(x,y,z)$.
Now you can calculate $xS_y\psi$ and $S_yx\psi$ and
see you get the same result.
$$\begin{align}
xS_y\psi
&= x \frac 12 \hbar \begin{pmatrix}0&-i\\i&0\end{pmatrix}
\begin{pmatrix}\psi_+(x,y,z)\\\psi_-(x,y,z)\end{pmatrix} \\
&= x \frac 12 \hbar
\begin{pmatrix}-i\psi_-(x,y,z)\\i\psi_+(x,y,z)\end{pmatrix} \\
&= \frac 12 \hbar
\begin{pmatrix}-ix\psi_-(x,y,z)\\ix\psi_+(x,y,z)\end{pmatrix}
\end{align}$$
and
$$\begin{align}
S_yx\psi
&= \frac 12 \hbar \begin{pmatrix}0&-i\\i&0\end{pmatrix}
x\begin{pmatrix}\psi_+(x,y,z)\\\psi_-(x,y,z)\end{pmatrix} \\
&= \frac 12 \hbar \begin{pmatrix}0&-i\\i&0\end{pmatrix}
\begin{pmatrix}x\psi_+(x,y,z)\\x\psi_-(x,y,z)\end{pmatrix} \\
&= \frac 12 \hbar
\begin{pmatrix}-ix\psi_-(x,y,z)\\ix\psi_+(x,y,z)\end{pmatrix}
\end{align}$$
So it is $xS_y\psi=S_yx\psi$ for every $\psi$, and hence $[x,S_y]=0$.