$Electrons do not emit or absorb photons: this would violate conservation of momentum in the electron’s rest frame. Electrons can scatter from photons. And if an atom emits a photon, it is common to say “the electron has changed states from $\ket A$ to $\ket B$” rather than “the atom has changed changed state.” Putting all of the change on the electron is roughly like assuming the nucleus is infinitely massive — but that assumption is better in heavy atoms, where you have to talk about the evolution of the entire multi-electron wavefunction in any case.
Your final paragraph (v1) asks about “virtual photons,” which are a computational tool in relativistic quantum mechanics, or quantum field theory (QFT). Virtual photons aren’t needed to understand entanglement. You might be amused, however, to learn that one model of atomic emission in QFT is that virtual photons are “scattered” from the charged constituents of the atom into real photons.
As for your actual questions:
I've been wondering how electromagnetic interactions between electrons are calculated if the position isn't determined until the wavefunction describing position collapses.
The position of a bound electron famously isn’t determined, apart from the overall shape of the wavefunction. While an individual atom may be well-localized within a solid, photon emission in solids tends to come from the delocalized electrons which make up the electronic energy-band structure. One example is the ensemble of light-emitting diodes which are probably illuminating this text on your computer screen.
Does the wavefunction need to collapse in order to emit a photon? Otherwise the photon's path would be undetermined
It’s perfectly fine for a photon’s path to be undetermined. A historical non-photon example. It’s harder for a photon’s energy to be undetermined, so there may be a “collapse” of the atomic wavefunction. But then again, there may not. In laser emission, the photon number operator doesn’t commute with the Hamiltonian which means you can’t say “there are exactly $N$ photons in this laser cavity.” I suspect (but haven’t verified) this also means you can’t determine the number of ground-state versus excited-state atoms in your laser medium.
It’s totally reasonable to imagine a matter interferometer where a single excited atom takes both of two paths, and in which light emitted by those atoms would originate from both paths, even in the single-photon limit. I’m not sure whether that has been done or not; most of the gee-whiz interferometer experiments are done with photons, because making coherent photons is really easy.