Consider an electron-positron scattering/annihilation. This process can also take place via gravitons (as a propagator) because fermions have mass. But the graviton has spin angular momentum of 2 units while fermions have 1/2.
So does conservation of spin angular momentum fail at the two fermion and graviton vertex. Even total angular momentum conservation fails... Why?
No, total angular momentum conservation does not fail. Spin and orbital such are not separately conserved.
Recall the graviton is defined as the lowest linear perturbation of the metric around a flat (Minkowski) background, $$ g_{\mu\nu}= \eta_{\mu\nu}+ \kappa h_{\mu\nu}. $$ The metric couples to the symmetrized energy-momentum tensor, (not just its mass term as you seem to think!), $$ \Theta^{\mu\nu}= \frac{i}{2}\overline \psi (\gamma^\mu \partial^\nu + \gamma^\nu \partial^\mu )\psi- g^{\mu\nu}\overline \psi (i\partial\!\! /-m)\psi . $$
As a result, a graviton couples to a fermion and an antifermion, analogously to the photon. (A standard SM calculation argues how the ensuing triangle anomalies cancel.) You might have to write the full messy amps down to detail how the total angular momentum is preserved, in a conspiracy between spins and orbital angular momentum. See Veltman—actually, the derivatives should be gravity covariant derivatives anchoring to the tangent space by means of Vierbeine, shrugged off for practical purposes here...
If you did your calculation with helicity, recall helicity is rotationally, but not Lorentz (!), invariant, and the graviton cannot have a rest frame, as it is massless. Note a three-point process for real particles is kinematically unattainable. The standard go-to review by Donoghue 1994 ArXive bypasses the tangent space peculiarities of fermion couplings, not central here.
While it is typically more important that conserved quantities are the same between the initial and final states of an entire process in QFT, it is certainly true that vertices, since they are typically related to terms in the Lagrangian, usually display the conservation properties as well.
It should be clear that $e^+e^-\rightarrow gr\ gr$ has spin configurations which conserve angular momentum since the allowed helicity states are $\pm\frac{1}{2}$ and $\pm 2$ for the fermions and gravitons respectively. The vertex however is more tricky. While I am no expert in low energy effective theories of quantum gravity, vertices with two fermions and a graviton do appear in papers on the subject, so there is likely some sort of solution to this conundrum. My educated guess is that since the quantity that must be conserved at the vertex is the total angular momentum, not the helicity itself, you are likely to see some restriction on the momenta of the particles involved. This type of restriction is quite normal in QFT as all vertices come with a momentum conserving delta function that guarantees linear momentum conservation throughout any diagram. An example of this is in particle decay wherein the final products must have equal and opposite momenta in the rest frame of the decaying particle.
My guess seems to align with the literature (some papers cited below) where researchers use symmetries, such as angular momentum, to restrict the number of form factors (general functions which typically tell you how the momenta involved in a process couple to each other or external sources) or diagrams contributing to processes involving fermions and gravitons. It should be noted that whether the gravitons are on-shell or not plays a very important roll in many of the cases studied because massive spin-2 particles have five helicity states, not just two.
So, in summary, remember that spin is a type of angular momentum, not all of it, and that just because it seems like spin is not conserved in an interaction does not mean that angular momentum is not; we will just get a restriction on the allowed correlations of the momenta involved.
Scattering of Fermions by Gravitons
Interaction of Dirac and Majorana Neutrinos with Weak Gravitational Fields
The end product will necessarily have at least two gravitons. Compare to QED where electron-positron annihilation produces at least two photons.
Without proof or reference, in general, at any order of perturbation, the sum of perturbation terms (diagrams), but not necessarily the individual terms, must reflect the symmetry of the underlying lagrangians.
