Your first question- Why aren't electrons being attracted by the positive charge region?
Any free charge will move in response to an electric field created by some charge distribution. So it's important to see the electric field in the region.
Well, the first thing you should do is find out the where the electric fields exist and where they don't. Electric field exists only in the depletion region, not in the 'neutral' p and n regions.
But then you ask why? The answer is clear if you know Gauss's law. It says the divergence of electric field is proportional to the net charge density. In the neutral regions there are no net (uncompensated) charges (Is a p-type neutral or charged? It's neutral. Because of each hole there's an acceptor ion that is negatively charged). On the other hand in the depletion region there are uncompensated positive and negative charges, and this creates the electric field.
To your second question, again, invoke the electric field: When you apply a voltage that increases the built in field, then you must have a larger number of uncompensated charges to support that field. That's why it expands in reverse bias and thins in forward bias.
In reality, you see, at equilibrium, electron and hole movement DOESN'T cease. There are still some electrons which are diffusing to the p-side, but an equal number is coming to the n side because of drift---electrons 'near' the depletion region on the p-side being sweeped by the electric field to the n-side. Similar things hold for holes.
Try reading Semiconductor Fundamentals by R F Pierret. It's a very good book for beginners.
I am struggling to understand the idea of holes being able to move across the junction, however.
Holes move in just the same way electrons do.
One minor difference is that because the curvature of the E-k curve for the valence band is different from the curvature for the conduction band, holes and electrons have different effective masses.
Other than that, they move the same way. If electrons can move across the junction, holes can do the same thing.
In comments you added,
As I understand it, the driving force behind electron diffusion is the repulsion of like charges. The holes are all in the valence band and so unlike electrons cannot move freely unaided, and thus I wouldn't expect them to naturally repel and move into the n-type region.
This is incorrect on a couple of points.
First, just like when uncharged molecules diffuse in a gas, the driving force behind diffusion is nothing more than that the particles are moving randomly, so if there are more of them at point A and fewer at a nearby point B, there will be a net flow of particles from A towards B.
To the extent that the mutual repulsion of the electrons causes a current, we would call that part of the drift current, not part of the diffusion current. (Also remember that, for example, in the n-region there are vast numbers of electrons and they repel each other, there are also fixed positive charges at the ionized donor sites that attract the mobile electrons, and the associated fields can cancel each other out)
Second, electrons can't move freely in the valence band because nearly all electron states are full. But this doesn't prevent holes from moving in the valence band. Holes in the valence band move readily, because the valence band isn't jam-packed with holes, it's jam-packed with electrons. Holes in the valence band move nearly as readily as electrons in the conduction band (see the point about different effective masses above).
In your comment reply to boyfarrel, you said,
Would it be correct to say that the diffusion of the free electrons from the n-type region into the p-type region causes electrons from thermally generated pairs to do the same, thus causing holes to gradually progress further away from the p-type region (and so into the n-type)?
This is again not quite right.
There are (relatively) many electrons in the n-region. These not only got excited from the donor sites into the conduction band, but thermodynamics says they'll take a certain distribution of energies above the conduction band edge. Because of this, a small fraction of them will have enough energy to overcome the potential barrier of the junction, should they randomly happen to move in that direction.
Similarly, on the p-side there are relatively many holes, and a small fraction of these have sufficient energy to overcome the junction barrier should they randomly happen to move in that direction.
The small fraction of the holes from the p-side that happen to cross the barrier over to the n-side form an excess population relative to the equilibrium hole population on the n-side. So, through random motion they have net motion away from the junction.
At the same time, as they randomly move around on the n-side, they have a probability to recombine with the vastly more numerous electrons present on the n-side. So we see (if the n-side is big enough) an exponential drop in this excess hole population as we move away from the junction. (the same way there's an excess electron population on the p-side, dropping exponentially as we move further into the p-side)
The p-side excess electrons aren't there because of anything the holes did, and the n-side excess holes aren't there because of anything the electrons did. They're there because there's a big honking crowd of electrons or holes on the other side of the junction, and a small fraction of those have managed to "jump the fence" and end up on the "wrong" side of the junction.
Best Answer
It is only possible for the donor and acceptor atoms to de-ionise in the depletion region if they capture a free carrier (electron and hole, respectively). But there are no free carriers in the depletion region because they have all be swept out by the strong electric field (something like 30$-$40kV/cm$^{\textrm{-1}}$!).
The short answer is because the carrier trapped by the dopant atoms would have to gain almost a bandgaps worth of energy to de-ionize.
The longer answer. Let's assume an acceptor atom in the p-side depletion region de-ionises by giving up it's captured electron. What happens? The electron is pushed back to the n-side by the field. However, the system is now no longer in equilibrium because the p-side is charged to +1 and the n-side is charged to -1. This is not stable! You can see that if you run this forward in time, eventually an electron from the n-side will have to neutralise the acceptor, bringing the material back to charge neutrality.
When you solve the Poisson equation for the pn-junction this is what you are solving for: the equilibrium distribution for charge neutrality. There are probably carrier dynamics like de-ionisation happening but they only serve to push the system out of equilibrium temporarily, eventually equilibrium will always be restored.