first of all, the quantum mechanical description 1 in its current form is impossible. A charged particle can't simply absorb a real photon. This is most easily seen in the final particle's rest frame. There is no photon left so the total energy is just the rest mass of the particle, times $c^2$, but the initial state has a higher energy of the particle in this frame, because the particle was moving, and also an additional positive energy of the photon. So the energy couldn't have been conserved in this process.
A charged particle may only absorb a real photon if it emits another one in another direction. So microscopically, it's always a process composed out of the Compton scattering subprocesses. If the number of real photons in the same state is large, they may be described as a classical wave. You may partially quantize the system, keeping the electromagnetic field classical and quantizing just the particle, or otherwise. The reason why this description agrees with the full quantum description in the classical limit is manifest.
Also, I don't quite follow the difference between your classical description and the quantum mechanical description 2. Concerning the usage of virtual photons, well, if you want to study the whole process using the tools of quantum field theory - including the pre-history when no electromagnetic wave existed - then of course the electromagnetic wave had to be created at some point, and some of the photons were absorbed by the charged particle. The charged particle became virtual for a little moment as well, before it emitted another photon which is needed for the energy conservation, as I have explained.
So the photon that was absorbed by the charged particle was virtual - it only existed for a finite amount of time. However, the electromagnetic wave was probably propagating for such a long time that even this photon may be called "real". There is a simple relationship between virtual and real particles - real particles are the virtual ones that happen to sit exactly on the mass shell, so they satisfy $E^2-p^2=m^2$. This identity may be exactly checked only when $E,p$ are measured totally accurately - which means that the particles must exist indefinitely. If they don't exist indefinitely, then they're always "virtual" to some extent, but chances are that if they exist for a long time, you may also imagine that they're "real".
The "virtuality" of a particle may be defined as the difference $E^2-p^2-m^2$ - the distance from the physical mass shell. If the virtuality is low, the virtual particle may exist for a long time and look "real".
Finally, there are no "virtual photon states" in the Hilbert space. The Hilbert space only contains real particles. Virtual particles are an object that appears in the calculation of probability amplitudes for various processes - in the Feynman diagrams. Virtual particles are internal lines of Feynman diagrams, given by propagators that determine the 2-point function (correlator) of a quantum field. But they don't correspond to any physical states. There are no off-shell physical states in the Hilbert space.
So if you have a history in which some particles exist for a finite amount of time, so that they're strictly speaking virtual from the Feynman-diagrammatic viewpoint, it is still true that at every moment, there must exist some real particles that are actually present. However, it is tough, misleading, ambiguous, and unnecessary to calculate the "exact intermediate states" in quantum field theory. Such objects - wave functionals - would also depend on the field redefinitions (of the quantum fields), renormalization schemes, and other things. It's actually very useful to avoid these things when they're not necessary and only talk about the things that can be measured - the cross sections that may be calculated from the scattering amplitudes.
A problem with the "wave functionals" of the intermediate states is that they're only well-defined with respect to a reference frame - but virtually all regularizations we know to calculate the loop diagrams rely on the Lorentz symmetry. Because the Lorentz symmetry is obscure by the foliations of the spacetime, it becomes harder to "regulate" the exact wave functional at the loop level. Of course, at the classical or semiclassical level, one may describe very accurately what's going on.
In your particular situation, there was no real problem because all the photons in the problem were really on-shell, and you may present them as real photons if you wish.
Best wishes
Lubos
Best Answer
In our modern understanding, every electron is thought to be a localized excitation of the electron (or Dirac) (spinor) field $\Psi(x^\mu)$, while every photon is considered to be an excitation of the photon (vector) field $A^\nu(x^\mu)$, which is the quantum field-theoretic counterpart of the classical four-potential.
Thus, the answer to your questions are:
All particles of the same type (e.g. photons or electrons) is understood to be 'coming from' one all-permeating quantum field. It should be noted that these fields also give rise to the corresponding anti-particles, so the positron field is the same as the electron field.
The different particle types are truly separated in quantum field theory: Each type is represented by one field, and the fields interact. These interactions are quantified by the Lagrangian (density), which essentially determines everything about the theory. In pure electrodynamics, the quantum field-theoretic Lagrangian density is (using 'mostly minus' sign convention for the metric)
$$\mathcal{L}_{\text{QED}}= \bar\Psi(i\gamma^\mu D_\mu-m)\Psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} =\bar\Psi(i\gamma^\mu (\partial_\mu+ieA_\mu)-m)\Psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} $$ where $F_{\mu\nu}\equiv \partial_\mu A_\nu-\partial_\nu A_\mu$ is the electromagnetic field strength tensor. The 'covariant derivative' $D_\mu\equiv \partial_\mu+ie A_\mu$ encodes the interaction between the two fields $A_\mu$ and $\Psi$, and the 'strength' of the interaction is given by $e$, the charge of the electron.