In LED's do the number of charge carriers (electrons and holes) decrease with time? According to page-1268-69 of Halliday, Walker & Resnick's Fundamentals of Physics (10th edition),

To emit enough light to be useful as an LED, the material must have a
suitably large number of electron-hole transitions....What we need is
a semiconductor material with a very large number of electrons in the
conduction band and a correspondingly large number of holes in the
valence band. A device with this property can be fabricated by placing
a strong forward bias on a heavily doped p-n junction, as in Fig.
41-16. In such an arrangement the current I through the device serves
to inject electrons into the n-type material and to inject holes into
the p-type material. If the doping is heavy enough and the current is
great enough, the depletion zone can become very narrow, perhaps only
a few micrometers wide. The result is a great number density of
electrons in the n-type material facing a correspondingly great number
density of holes in the p-type material, across the narrow depletion
zone. With such great number of densities so near each other, many
electron-hole combinations occur, causing light to be emitted from
that zone.

Now, if electron-hole pairs are ceasing to exist due to recombination and resulting in a greater number of "gridlocked" electrons and light, who will continue to conduct electricity? Won't current flow stop after a while due to the absence of electron-hole pairs?
 A: You won't run out of electrons and holes. There are two main processes through which you will get more.
Thermal energy will naturally generate electron/hole pairs in your semiconductor. That's where they come from in a semiconductor at thermal equilibrium. However, this process is slower than we would like for an LED, or any diode. Fortunately, we can help it along by injecting carriers from outside the p-n junction. We can imagine a simple p-n junction with metal contacts set back a bit from the p-n junction. You can directly inject electrons into the n side, replenishing the lost electrons. At the p side you will have a greatly increased electron/hole generation rate allowing you to effectively inject holes into the p-side (by generating an electron/hole pair and removing the electron into the metal).
A: No, the current won't stop flowing as the battery maintains constant electromotive force. Some free electrons will lose their kinetic energy and become gridlocked. You can view this phenomenon, that is, free electrons losing their kinetic energy, as free electrons losing kinetic energy to a resistance, say a light bulb.
