What is the generalization of the path integral to the case of particle interaction?

Yes.

In something like:

$$ e^-+mu^-rightarrow e^-+mu^- $$

you start with two free particle states, and end with 2 free particle states. In between, the electron and electromagnetic fields (and perhaps the weak field) go through all possible configurations that conserve energy an momentum. That is something that cannot be solved exactly, so it is approximated with Feynman diagrams:

Note that this only includes connected diagrams. In theory you need to consider a field configuration with 2 separate diagrams, say diagram (a) plus a photon loop off to the side, in practice they are included by exponentiating the sum of connected diagrams. (How that works, is technical).

Also, if you look at the higher order diagrams, say (d), there can be an arbitrary four momentum running around that electron-position loop in the photon line, so it in fact represents an infinite number of field configurations, but since you just add diagrams, that's what the integral does.

Each electron-photon vertex does include a factor of $-i(-e)$, but that's not the only contribution to the diagram. You simply have to evaluate them to find the total contribution. (A positron-photon vertex has $-i(+e)$, which may not make sense if you watch youtube videos where attraction is described as exchanging a boomerang between 2 row boats, while repulsion is a basketball...that is simply putting too much classical reality into virtual particle exchange).

Several things to note:

In the tree level diagram ($t$-channel), the exchanged photon is highly virtual. For instance, in the center of mass frame, with incoming energy $E gg m_{mu}$, and scattering at 180 degrees, it's 4-momentum is:

$$ q^{mu} = p^{mu}-p'^{mu} = (E,0,0,E)-(E,0,0,-E) = (0,0,0,2E)$$

That is, it has zero energy and lots of momentum. As a particle, it is space-like (hence:virtual). It is not emitted by the muon and the absorbed by the electron, nor is it emitted by muon and then absorbed by the electron. It's just "exchanged".

Meanwhile, in $s$-channel $e^-e^+$-annihilation, you find:

$$ q^{mu} = p_1^{mu}+p_2^{mu} = (E,0,0,E)+(E,0,0,-E) = (2E,0,0,0)$$

and the photon has lots of mass and zero momentum. Again: not a property of real photons.

Nevertheless, virtual photons have longitudinal and transverse polarizations (depending only on the kinematics of the external lines), which can be used to separate the electric and magnetic responses of nuclear targets at fixed momentum transfer. It makes them seem quite like real particles, but they are indeed virtual.

For identical particles in the final state, you need to include $u$-channel processes:

that swap the final state four-momenta.

Finally, all these diagrams are amplitudes. They interfere. At tree-level, the exchanged particle could be a photon, and it could be a $Z^0$, hence both processes contribute, and their interference term is parity violating. (This fact is used to probe the spin of the strange quark sea via elastic scattering of polarized electrons off protons).

The point is: all diagrams contribute, and you have to add them all up to get the amplitude. Only then do you square it to get a probability, which is converted into a differential cross-section to compare to experiment.

Moreover, virtual particles are perturbative approximations to intermediate field configurations, that's it. They are not actually particles, but it can useful to treat them as such in oder to interpret scattering experiments.