I don't believe you question is well formed. Note that the emission or absorption of a single photon by a charged particle is just a single term in one choice of a perturbative expansion of a total physical interaction. Take a classic (not classical [edit]) electron-electron elastic scattering in the center of mass picture.
Here comes Electron A in from the left and Electron B in from the right. Then later there goes A at some angle and B at the opposite angle. Total kinetic energy unchanged, total momentum still 0. But the momentum of each electron changed due to the electromagnetic field and therefore there has been an exchange of photons. How many and which way? That's indeterminate and complementary to the observed behavior presupposed here. There was some superposition of 0, 1, 2,... photons exchanged subject only to the constraint that the later observed deflection yielded the given momentum exchange on the electrons.
Now when you consider such a collision classically you also know that the mutual acceleration of the two charged electrons will induce a classical electromagnetic wave. To account quantum mechanically for this you must treat the formerly elastic collision of two bodies as a many body inelastic collision. This correction, a "collision" of the two electrons plus the quantum e-m field is the Bremsstrahlung radiation. There will be a net emission of a superposition of 0 or 1 or 2 or .... photons of various frequencies.
How much on average is calculated via field theory using perturbative expansions in which the picture drawn of one electron emitting a photon is just a single term in the sum-over-histories calculation.
As to "inverse Brem.." in the presence of a thermal background photon gas there is always the probability that two scattering electrons come out with more energy then they had prior to scattering but that probability is very low. Entropy increase dictates that the net effect is a thermalization of the energy equipartitioning between electron motion and the thermal photonic environment.