We have virtual particles in quantum field theory (QFT). In general, they don't have the need to obey causality.
My question is:
Do the processes in QFT (electron self-energy, photon self-energy, electron-photon vertex, etc.) have to obey causality?
For example, can some parts of the electron self-energy diagram fall out of the light cone?
Or can some parts of the electron-positron annihilation process fall out of the light cone?
Quantum fields are fields that have observables at each spacetime point. In QFT these observables evolve according to local, relativistic equations of motion. In addition, the observables at any two spacelike separated points commute with one another.
A particle is a particular kind of state of a quantum field that acts in nice particle like ways. If you try to measure the number of particles on a level smaller than the Compton wavelength $h/mc$, which is the wavelength of a photon whose energy is the same as the rest mass of the particle, that requires interacting with other fields with energy and momentum components on the scale of the inverse Compton wavelength. This can produce particle anti-particle pairs of the particle you're tying to localise and this undoes your attempt at localisation.
Since real a particle states isn't localised at a particular point but is instead a bit fuzzy, so is its "light cone". The notion that a particle is going outside of its light cone is based on an unphysical idea that it was localised at a particular point.
In "Quantum field theory for the gifted amateur" by Lancaster and Blundell, section 8.3 they do a calculation of the amplitude for a relativistic particle propagating from one point to another. they find that it leads to the particle propagating outside its light cone. This sort of calculation can cause confusion because there is no such thing as a particle being at a single point in QFT. Lancaster and Blundell try to treat this as a reductio ad absurdum of single particle relativistic quantum mechanics and mention the Compton wavelength issue at the end of the section, but this seems to confuse a lot of readers.
A better discussion of locality in quantum field theory can be found in "The conceptual framework of quantum field theory" by Anthony Duncan, see especially Chapter 6, Section 6.5.
Regarding your first question: Every observable process have to obey causality. One of the reason why QFT is created in the first place is that we want a quantum theory obeying the causality.
This point has been clarified well enough by the previous answers. But looking through the question it seems that OP's question is more concentrated on the causality in Feynman diagram.
It is important not to mistake Feynman diagram for an actual process. Feynman diagram is just a way Feynman introduce to simplify the calculation of perturbation in QFT, so the process you see in a Feynman diagram, like one electron and positron collide and generate a virtual photon (mind the word "virtual") is not actually happening. And that' s why virtual particles might violate causality--they don't actually exist.
No, to my knowledge causality is a crucial part of quantum field theory. Everything that can be measured has to obey causality. For bosons, this corresponds to the fact that $\langle 0|[\phi(x), \phi^*(y)]|0 \rangle$ is always zero at space-like separation. For fermions, we know that $\langle 0| \lbrace \psi(x), \bar{\psi}(y) \rbrace |0 \rangle$ is zero outside the light cone. Since operators such as charge, energy, and momentum always involve an even number of spinor fields, this is enough to ensure that $[\mathcal{O}_1(x), \mathcal{O}_2(y)]$ is also zero outside the light-cone. Which says no two measurements can effect each other at space-like separation.
By the way, I don't have a mathmatical proof of how the anticommutation relation is related to the commutator of the operators. I would love to see one.
