A magnetic dipole transition can be modelled as a time-dependent perturbation $V_{\text{md}}(t) = {e\over 2 m}(\vec{L} + 2\vec{S})\cdot \vec{B}e^{-i \omega t}$. Fermi's Golden Rule tells us that the transition rate for $b-X,1$ is proportional to the matrix element of the perturbation between the initial and final states,
$$W \propto \langle \psi_b|{e\over 2 m}(\vec{L} + 2\vec{S})\cdot \vec{B}|\psi_{X}\rangle,$$
where $|\psi_b\rangle$ is the excited state and $|\psi_{X}\rangle$ is the ground state (with three possible $M_S$ values.)
The effect of $\vec L$ and $\vec S$ will be to turn the final state into some combination of the triplet states, but it won't change $J$. Therefore we might expect the transition to be 'spin-forbidden':
$$W \propto \langle b^1\Sigma_g^+ |X^3\Sigma_{g,M_S=0,\pm1}^-\rangle = \langle J=0 | J=1 \rangle = 0.$$
This is where the spin-orbit coupling comes into play. Spin-orbit coupling is the reason why the singlet $(b)$ state has a higher energy than the triplet $(X)$ states. It is a perturbation of the form $V_\text{SO}={\mu\over\hbar}\vec{L}\cdot\vec{S}$, which can be rewritten as $V_\text{SO}={\mu \over 2\hbar}(J^2-L^2-S^2)$. In a spherically symmetric system like the helium atom, this perturbation commutes with the Hamiltonian, so all you get is a shift in the energy of the triplet (L=1) and singlet (L=0) states. However, in a linear molecule like $O_2$ you lose the spherical symmetry, so $[L^2,H]\neq0$ and in addition to an energy shift, you also get some mixing of the unperturbed eigenstates, so that the excited state is not exactly $|b^1\Sigma_g^+\rangle$, but rather $|\psi_b\rangle = c_1|b^1\Sigma_g^+\rangle + c_2|X^3\Sigma_{g,M_S=0}^-\rangle$. This mixing of J=0 and J=1 states is what allows $W$ to have a nonzero value. Since we can write $S_x = S_+ + S_-$, there will be a term in the transition rate like
$$W\propto c_2^*\langle X^3\Sigma_{g,M_S=0}^-|S_{\pm}|X^3\Sigma_{g,M_S=\mp1}^-\rangle+\cdots \neq 0.$$
Does this help? I know this is a bit hand-wavy so let me know if I can clarify anything.
The selection rule $\Delta S=0$ is an approximation, nothing more, and in suitable circumstances it can easily break. One prominent example of this is the hydrogen 21cm line.
Electromagnetic atomic and molecular transitions are arranged into a series in order of multipolarity, which describes the atomic operators that enact the interaction hamiltonian, with higher multipolarities scaling down in coupling strength as powers of $a/\lambda$, i.e. the ratio between the system's size and the radiation's wavelength, which is generally small. Thus, you get
- as the leading-order term, electric dipole (E1) transitions;
- weaker by a factor of ${\sim}a/\lambda$,
- magnetic dipole (M1), and
- electric quadrupole (E2) transitions;
- still weaker by another factor of ${\sim}a/\lambda$,
- magnetic quadrupole (M2), and
- electric octupole (E3) transitions;
- and so on.
Generally, the no-spin-flips rule holds only for electric dipole transitions, for which the interaction hamiltonian is the electric dipole operator $\hat{\mathbf d}$, which does not couple sectors with different spin (unless you have strong spin-orbit coupling).
However, it is perfectly possible to have magnetic dipole transitions between states of spin that differ by $\Delta S=1$. Here the coupling is weaker (so you will need a higher intensity, or longer pulse times, to excite them), and thus typically the linewidth is smaller (so you will require a sharper laser), also giving a longer decay lifetime, but those are things that make the transition harder to observe, not impossible.
Best Answer
The selection rule for $m$ corresponds to the conservation of the $J_z$ component of the angular momentum, where $z$ is the chosen quantization axis.
The $\Delta m=0$ possibility occurs when the system is driven by light which is linearly polarized along the quantization axis, in which case each photon has total angular momentum $1$ but nevertheless has $J_z=0$.
The $\Delta m=\pm1$ case happens when the system is driven by light which is circularly polarized in the plane orthogonal to the quantization axis, in which case each photon has angular momentum component $J_z=\pm 1$ that must be absorbed by the system.