The lower part is not filled with quasi-particles. At zero Kelvin, in zero magnetic field and with zero disorder all free electrons condense and form the superconducting condensate. The semiconductor model now describes the breaking of Cooper pairs not as resulting in two electron-, but in one electron- and hole-like excitation. As you are potentially familiar with in semiconductors, one electron of energy $E$ above the Fermi level may be described by one hole with the same energy below.
Now, this is the equilibrium. Tunnelling processes however are very non-equilibrium processes! When you apply a bias across your junction you give rise to quasi-particle excitations, with result in a tunnelling current. That is why, even at 0 Kelvin, there is "electron" tunnelling...
First of all, at zero temperature not all of the electrons in the BCS ground sate (GS) form Cooper pairs. One way to think about this is that deep inside the Fermi surface, there is no available states for scattering events, including the phonon-mediated ones, and thus we wouldn't have the effective attraction to form the Cooper pairs.
This can also be seen from the coefficients $u_k$ and $v_k$ in the postulated BCS GS given in the original post. When $k$ is way below the Fermi surface, then $|\xi_k|>>\Delta$, with $\xi_k<0$. In this case, $|u_k|\to 0$ and $v_k\to 1$, i.e. the states far below the Fermi surface are almost fully occupied by electrons created by $c^{\dagger}_{k\uparrow}c^{\dagger}_{-k\downarrow}$. These electrons are created in pairs, but they are NOT Cooper pairs!!! The pair creation operator does not tell us anything regarding whether the pair is a Cooper or just a normal pair. Only when it is close to the Fermi surface, and thus possible to have phonon-mediated scattering, is the Cooper pair creation possible.
In summary, the BCS GS looks like the following: deep down inside the Fermi surface, it's just normal electrons. Close to the Fermi surface, it's mostly a superposition $\left(u_k+v_kc^{\dagger}_{k\uparrow}c^{\dagger}_{-k\downarrow}\right)|0\rangle$ of empty state, with amplitude $u_k$, and occupied Cooper pairs, with amplitude $v_k$.
EDIT:
After I submitted the above answer, I realized that I missed out the other part of your question, which is about the excitations of the BCS GS and here it goes:
You mentioned that "the excitations of the BCS state must be created or destroyed in pairs". This is not correct. The excitations of the BCS GS are fundamentally different from Cooper pairs or the normal electrons. Instead, it is the quasi-particle created by
$$\gamma^{\dagger}_{p\uparrow}=u^*_pc^{\dagger}_{p\uparrow}-v^*_pc_{-p\downarrow}$$
Not surprisingly, this is just the Bogolyubov transformation used to diagonalize the BCS mean field hamiltonian and this why it is the legitimate excitations of the GS. The corresponding eigenvalue is exactly
$$E_k=\sqrt{\xi_k^2+\Delta^2}$$
This is where the gap $\Delta$ to the BCS excitation comes from. This kind of excitation is slightly less intuitive compared to objects like single electrons or Cooper pairs. It has spin-1/2, but it doesn't carry well-defined charge, whereas a Cooper pair has spin 0 and charge $2e$.
Also, You can easily check that
$$\gamma_{p\uparrow}|BCS\rangle=\gamma_{-p\downarrow}|BCS\rangle=0$$
which makes perfect sense since the annihilation operator for the excitation annihilates the GS.
Best Answer
The density of states shown is the single electron density of states. In other words, it shows the density of single electron states. Once the electrons condense into Cooper pairs, they cannot be described with single electron states, and thus "vanish" out of your single electron state description. Assuming zero temperature, all states to the left of $\epsilon_F$ are filled (valence band), all states to the right are empty (conduction band). Current is now carried by Cooper pairs, not single electrons.
To break a Cooper pair, you need to place each electron in the conduction band, which requires about $2 \Delta$ energy. These electrons were not in any other single electron states of the DoS before.
On the bands: At zero temperature, all states below $\epsilon_F$ are filled, and all states above are empty. In a metal, $\epsilon_F$ is in the conduction band, and moving $\epsilon_F$ slightly up or down changes how many electron states are filled. In an insulator, $\epsilon_F$ is in the gap between the bands, and moving $\epsilon_F$ slightly changes nothing. In terms of bands, it makes sense to consider the superconductor an insulator, since $\epsilon_F$ is in a gap, and moving it does not change anything for the electron states.
In a sense, the superconductor is an electron insulator, since no electrons can carry current, only Cooper pairs. By carving a gap through the conduction band, we have turned the metal into a kind of insulator, and the bottom part of the previous conductance band is now a filled valence band.
This is all in terms of electron states, not the electrons themselves. Trying to develop an intuition for which electrons go into the condensate is a futile task. All of them went into the condensate, and the DoS diagram no longer shows where the electrons are, but where they can go if we excite the superconductor out of its ground state.