Do electrons always emit photons?

You, like many others, have fallen victim to a rude and inaccurate analogy. Such analogies are always misleading, they cause more questions than they give answers.

To you, your question seems completely logical to ask, but to a trained physicist it is a meaningless question.

The only way to truly understand the physics of electron-photon interaction is to study the mathematics of Quantum Electrodynamics, one of the simplest models of interacting Quantum Field Theory. There are no royal roads to science. Do not trust analogies, even when repeated by senior researchers. They think they are making the subject more approachable to the general public, but it mostly results in widespread confusion and misconception.

The photon that electrons "use to mediate interaction" is actually not at all a physical object, or anything of the sort. It is a mathematical label that is used to diagrammatically encode terms – complicated mathematical expressions – in the quantum field theoretical formula that gives you an approximate probability of observing an end state of the field (configuration of electrons with certain values of coordinate/momenta) given an initial state of the field. This quantity is called the S-matrix.

Unfortunately, I cannot possibly explain to you Quantum Field Theory in a single answer here, but I strongly encourage you to study it. Fortunately there are plenty of beginner friendly textbooks out there in the year 2020.

Based on various professional scientists'/teachers' youtube channel comments: virtual particles really captivate the non-expert and cause nothing but confusion.

Problem (1) is that the expert always explains them in terms of old-fashioned perturbation theory, in which the energy to exist for a short time is borrowed from the vacuum per the H.U.P. The virtual particles are time ordered, so one particle emits one and the other absorbs one. In beta decay, they always drop the 1/2 of the process where the d-quark absorbs a W+ that has already produced the final state leptons (and likewise in other processes)...they tell half the story.

In modern Lorentz covariant formulations, all diagrams conserve energy and momentum at each vertex, so the question "where does the energy come from" is basically meaningless.

People also ask how emitting/receiving a photon can cause attraction: again, 4-momentum is conserved at all vertices, so the photon has the correct 4 momentum in a scattering process. Then they ask: which particle emits and which receives? Well, Feynman diagrams with an exchange particle encompass both time-ordered diagrams in a covariant way: the question is again, meaningless in the context of the theory.

Then comes bound states and "continuous emission"...which may make sense in the context of self energy and renormalization, but that is total advanced.

The simple tree level diagrams are used in scattering theory. Here we have initial and final states approximated by free particle states, and the transition matrix encompasses all possible paths....that is not solvable, so perturbation theory is used, and that's where the diagrams come in.

The tree level diagram is an exchange of a single virtual particle. The particle is completely off shell. For instance, in s channel:

$$ e^++e^- rightarrow gamma rightarrow X $$

at a collider, the photon has zero momentum and lots of energy (it's basically a massive particle at rest in the lab...a photon?). Conversely, in scattering (t channel) in the Breit-frame, the virtual photon has zero energy and space-like momentum, so treating that as "real" can only cause confusion.

So the non-expert can waste a lot of time reading to much into virtual particles. At the tree level, they are completely defined by the kinematics of the initial and final states, and at loop level, you have to integrate of all 4-momenta, so anything goes.

Nevertheless, to the expert, they are very useful for understand physical processes, e.g. the Rosenbluth separation in $ep$ scattering: the longitudinal and transverse polarization of the virtual photon (again: defined by kinematics) can be used to separate the electric and magnetic responses of the proton at fixed $Q^2$.

In parity violating elastic $ep$ scattering, the entire signal can be viewed as interference between 1 photon exchange and 1 $Z^0$ exchange, which is very handy.

So, in summary, they are very useful for experts, and cause lots of confusion for non-experts, and most of the common non-expert questions fall into the "that's not what virtual particle are" category.