Why do the drift and diffusion components cancel for each type of carrier if EHP generation plays such big role in p-n-junctions? I have always argued to myself that drift and diffusion components of the current though a p-n-junction cancel for each type of carrier because any electron diffusing from n into p will sooner or later require the same electron to drift back. (And analogously for holes).
However, it now seems to me that the generation of EHPs (electron-hole-pairs) profoundly contributes to the drift current:

"The supply of minority carriers on each side of the junction required to
  participate in the drift component of current is generated by thermal excitation of electron-hole pairs." (In "Solid-State-Electronic Devices" , Streetman and Banerjee)

In thermal equilibrium, that implies an equal amount of recombination events.
Anyway, one can imagine an EHP being generated on the p-side, with the electron drifting and the hole diffusing to the n-side, where thy recombine. In that scenario, there is no hole drift and no electron diffusion, but hole diffusion and electron drift. 
So, how in earth does one dare to argue that "the drift and diffusion
components must cancel for each type of carrier"?
I feel like I am missing the real argument.
 A: The concentration of electrons ($n$) and holes ($p$) in a semiconductor is determined by the Fermi-Dirac distribution function. As you correctly mentioned, the Electron-Hole Pair (EHP) generation and recombination rates are equal in thermal equilibrium. As a result, when the semiconductor is in thermal equilibrium, the concentrations $n = n_0$ and $p = p_0$ are constant (not necessarily equal). If for some reason these quantities are (say) depleted (i.e. $n < n_0$ and $p < p_0$) then the rate of EHP generation will exceed the rate of recombination until $n$ and $p$ are back to $n_0$ and $p_0$ respectively. In other words, the system returns to thermal equilibrium.
Now, in your analysis you are forgetting that there is EHP generation occurring on the n-side as well. When a EHP is generated on the n-side, the hole drifts and electron diffuses to the p-side. The current due to the electrons drifting to the n-side must cancel the current due to electrons diffusing to the p-side. Similarly, the current due to the holes diffusing to the n-side must cancel the holes drifting to the p-side. If they did not cancel each other perfectly, there would be an accumulation or depletion of holes or electrons on the n- or p-side depending on which current is dominant. When I say accumulation or depletion, I mean deviation of $n$ and $p$ from $n_0$ and $p_0$ respectively. And as I mentioned in the previous paragraph, deviation of $n$ or $p$ from $n_0$ and $p_0$ will cause the EHP generation and recombination rates to adjust such that the system returns to $n_0$ and $p_0$.
