$rm SU(2)$ symmetry and rotation invariance

Let's break down the sentence "Invariance with respect to rotation of the electron’s spin, or SU(2) symmetry, leads to conservation of spin polarization". The invariance means that the Hamiltonian commutes with total spin, $[H,vec{S}] =0$. (You can show this also by requiring that $H$ is invariant under arbitrary spin rotation $U= e^{i vec{s}cdot vec{n} theta}$, $H=UHU^dagger$, see also the answer by Undead.) From this it follows that we can simultaneously diagonalize $H$ and any component of $vec{S}$. If we diagonalize for example $S_z$, you find that any state with $S_z$ eigenvalue $+1/2$ will keep that eigenvalue, even if it's not an eigenstate of $H$. An example of a Hamiltonian that does not conserve spin, is the spin-orbit term $(vec{E} times i hbar nabla)cdot vec{S} $ (I believe something like this is considered in the paper you cited). Here $vec{E}$ is some external electric field which intuitively breaks the rotational invariance of the system. As was mentioned by Undead in their answer, even this Hamiltonian has a symmetry: if you rotate both the spin and the angular momentum ($vec{E} times i hbar nabla$), you can keep this term invariant.

Formally, we consider that a quantum mechanical system is symmetric with respect to a continuous transformation if its Hamiltonian commutes with associated symmetry generator. In this case, the generator of the spin rotation transformation is simply the spin operator.

If the Hamiltonian commutes with the spin operator $$ [H,vec{S}] = 0 $$ and if $left | Omega right rangle $ is an eigenstate of the Hamiltonian (say the ground state) it implies that $$ e^{i vec{n} cdot vec{S}} left | Omega right rangle = e^{itheta} left | Omega right rangle $$ so that a spin rotation has no effect other than an unimportant phase factor and the state is invariant.

Spin conservation (Noether’s theorem) follows directly from the first equation. You can easily verify that spin-orbit coupling breaks spin rotation symmetry.

However, remember that experimentally a spin rotation is not as easy to generate as to just apply $e^{i vec{n} cdot vec{S}}$ on a state. What is common to do is a rotation of both angular momentums $J = L+S$ with magnetic fields. Spin-orbit coupling commutes with that, in other words, it conserves angular momentum which is expected.