Can a photon's angular momentum couple to an electron spin, to cause a spin-flip (thus allowing intersystem crossings)? Been reading up on intersystem crossing and have gotten rather confused by:

a photon possesses an angular momentum of $h/2\pi$ so it can couple to an
electron spin and in principle induce any plausible spin transition
between two spin states

This suggests to me that in a system with two coupled electron spins, the absorbed photon can provide angular momentum to allow a spin-flip between a singlet state to a triplet state (and conserve total angular momentum). That is, an S (0) to T (1) transition is allowed if coupled to a photon.
I know this is not the case. Experimentally, the S (0) to T (1) transition has no oscillator strength and is not observed in an absorption spectra (even in molecules with no heavy-atoms for spin-orbit coupling). What am I neglecting the prevents the angular momentum of an absorbed photon coupling with an electron spin to allow a change in the spin?
 A: 
I know this is not the case.

This is a correct statement in the electric dipole approximation: there is a selection rule that forbids transitions with nonzero spin change.
However, the interaction of light with atoms and molecules, in its full sense, comes in a hierarchy of selection rules, ordered by decreasing importance: electric dipole, then magnetic dipole and electric quadrupole, then magnetic quadrupole and electric octupole, and so on. The transitions higher up in this ladder are typically (much) weaker than the ones lower down, but they often allow for transitions that would otherwise be forbidden.
In other words, spin-flip transitions are weak (and often weak enough that they can be completely ignored) but it doesn't mean that they don't happen (so see e.g. this line for a specific example).
A: The electric dipole transition between a state of triplets $|1,\pm 1\rangle$ and a singlet $|0,0\rangle$ has the amplitude
\begin{equation} 
\langle \psi_{singlet}|e\mathbf{r}|\psi_{triplet} \rangle . 
 \end{equation}
But since $\mathbf{r}$ does not act on the hilbertspace of spins, the amplitude is zero.
I hope this answers your question
