Is this something due to change in Hamiltonian due to relativistic correction. i.e. $H$ being $$(\frac{e^2}{4\pi {\epsilon _0}}) \mathbf{S \cdot L}$$ instead of $$(\frac{e^2}{8\pi {\epsilon _0}}) \mathbf{S \cdot L}$$ like in Thomas Precession.
Ref :-Page 278 DJ Griffiths Introduction To Quantum Mechanics.
Why don't they commute?
You can work out the maths. It does not have anything to do with the factor but the form.
$$ \begin{align} [\mathbf L \cdot \mathbf S, \mathbf L_x] & = [L_xS_x+L_yS_y +L_zS_z, L_x] \\ & = S_x[L_x,L_x] + S_y[L_y,L_x] + S_z[L_z,L_x] \\ & = 0 - S_y (i \hbar L_z) + S_z (i \hbar) L_y \\ & = i\hbar (\mathbf L \times \mathbf S)_x\end{align}$$
Similarly, you can work out other two components. Clearly, Hamiltonian does not commute with $\mathbf L$. The same kind of calculation goes with $\mathbf S$ with $\mathbf L \rightarrow \mathbf S$.
The spin-orbit interaction implies the presence of a torque between the orbital and the spin moments. The $z$ components cannot be constants of motion since they are precessing. What we hence need is the total angular momentum $\mathbf J =\mathbf L+\mathbf S$.
