When you excite a semiconductor you typically generate lots of electrons and holes. When studied in detail they will have a rich dynamic, as there are several recombination paths, as you mention in your question. The question is always how likely is a certain process is to happen, i.e. how big is its transition matrix element.
The necessary requirement for an optical transition is that you have an electron and a hole in two states, which allow an optical (dipole) transition. For this the typical selection rules must apply, and another important point is the difference in momentum. In Si for example, the highest VB state (at the $\Gamma$ point) and the lowest CB state (close to the $X$ point) have a strong $k$ difference and thus direct optical transitions are not possible. In this case you will need the assistance of phonons to bridge the k difference for the emission of a photon. Another typical process is the build up of exciton states which then later on may perform an optical recombination.
Non radiative transitions
As you suggested it is indeed possible to recombine electrons and holes without emitting a photon. The possibly simplest process is the Auger process, where the energy of the electrons and holes transfers to a 3rd particle (electron or hole) which is taking over the energy and becomes excited itself. The problem with a direct phonon emission is that you have to match energy and momentum with the phonon you create. But the properties of phonons are given by their dispersion relation, so the phonon you need might not exist. For perfect semiconductors these direct phonon processes does not play an important role (at least as far as I know - counterexamples are welcome). When defects and impurities come into play this changes. There a several examples of non-raditive relaxations of excited defect states.
For your question if the electron has to return to its initial state again, there is no such thing as "exactly this electron" in QM. Particles are indistinguishable so the answer is no.