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.
How come these Ionized Atoms aren't Displacing Electrons to create more holes (In the Ionized P-Side) or Accepting more electrons (In the Positively Ionized Region on the N-Side)?
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}}$!).
So why then do the electron from the n-side stay with the acceptor atoms on the p-side once the junction has formed?
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.
Best Answer
You have misunderstood things. A diode as a whole is neutral but not the p and n regions across the depletion layer. The video that you've linked is partially technically incorrect, But the inaccuracy is altogether different from your doubt. It correctly says that n-type and p-type regions are charged at 3:56 with the following words:
Both the p and n sides are initially neutral before the diffusion current starts. After the diffusion current starts, the holes from p side starts moving towards the n side. These holes carry a positive charge. So these will make the n region positively charged. Similarly, the electron moves from the n side towards the p side. Since electrons are negatively charged particles, they will make the p region negatively charged. Following is a figure of an ideal step junction diode, taken from wikipedea:
It is clear that the Electric field is from n to p. Note that the diffusion occurs because of the abrupt variation in the density of charge carriers across the depletion layer.
Now comes the technical mistakes in that video.
There isn't any neutral region in the depletion region. there is a plane of 0 thickness which is electrically neutral, as is clear from the picture above.
The diffusion current never stops. It would stop only when the concentration gradient would have ceased to exist -- but this never happens. As soon as the diffusion current starts, opposite charges start to build on both the sides creating an Electric field. Since the material is semiconductor, it has charge carriers in its whole body at room temperature. Even if some charge carriers get combine to neutralise each other, due to temperature new electron-hole pairs will be generated. The misconception that might develop is that the potential difference will stop the diffusion current altogether. Diffusion current consists of the charge carriers moving at high speeds because of their thermal energies. So, the Electric field can't stop the diffusion current, then what does happen next, do the charges keep piling up on p and n sides. No! the drift current comes into picture. As you know the diffusion current is only due to majority carriers, and can't be stopped. but there are minority carriers too. the potential difference across the depletion layer causes the minority carriers to move and create the drift current. When the steady-state reaches, the drift current becomes equal to the diffusion current. The hole-drift current is given by $J_p= qp\mu_p E$ and the hole-diffusion current is given by $qD_p \frac{dp}{dx}$, The expresions for electron-drift and electron-diffusion currents are similar, with the subscript being $n$. If you want I can add references and mathematical proves to vindicate my assertions.