It's a reversible equilibrium reaction. Neutron <-> proton + electron + antinutrino + energy. The forward reaction is beta decay, and is favored by lone neutrons and by the neutrons in some atomic nuclei (i.e. radioactive isotopes that undergo beta decay). Presumably the half-life of a nucleous is determined by the degree of favorability of the beta decay reaction within that nucleus. The article already mentions that the reverse reaction (neutron formation) is favored under the conditions of a neutron star, since the pressure favors the side of the reaction with fewer particles. The gravitational potential energy released as the neutron star collapses supplies the energy required for the reverse reaction.
In simple terms, it does decay, then immediately recombines.
When a neutron decays it makes two charged particles. But inside a neutron star those two charged particles (proton and electron) have no room to separate, instead, because they attract each other they immediately recombine.
Normally they can not recombine because of the energy gap - they need some extra energy to become a neutron (which is why hydrogen, which is also a proton and an electron doesn't just turn into a neutron).
But inside a neutron star that energy is not lost, instead it's transfered to some other set of protons and electrons, giving them the energy they need to become neutrons.
I think there are actually two forces at play, according to the article: gravity and degeneracy pressure.
One the one hand, the neutrons are stable and not decomposing into protons, electrons, etc. by the extreme gravity. At a previous point in the star's lifecycle, it was comprised of normal atoms which contain electrons, protons, and neutrons. In atoms under typical conditions, there is a force which keeps the electrons from falling into the nucleus. Later in the star's life, as the star ran out of fuel, gravity overcame the force that normally keeps electrons from falling into the nucleus and so the protons and electrons fused to become neutrons. Hence, a neutron star.
If one were to release the gravitational pressure, the neutrons would become unstable and you'd get the decay back to protons, electrons, and the radiation that's mentioned. It would be a huge release of energy, as the article states.
The degeneracy pressure is the other force at play. It's what keeps the neutrons from collapsing onto one another and becoming a black hole. I think it's essentially the Pauli exclusion principle, but I could be wrong there.
So, the degeneracy pressure keeps them separated as individual neutrons, but gravity is what is stabilizing the neutrons themselves as neutrons.
If the star were large enough, gravity would overcome even the degeneracy pressure, resulting in a black hole.
Degeneracy pressure is why the whole thing is stable against gravitational collapse, but I don't think it's why individual neutrons don't decay. (I'd say what is if it weren't for the trivial obstacle that I don't know. Handwavily, I think it's often the case that large ensembles of particles can get into nice stable low-energy states for subtle quantum-mechanical reasons.)
I wish he hadn't glossed over this. What is it that keeps them stable in atoms?
Edit: Wikipedia isn't very helpful, I may just be too dumb to understand this paragraph:
"When bound inside of a nucleus, the instability of a single neutron to beta decay is balanced against the instability that would be acquired by the nucleus as a whole if an additional proton were to participate in repulsive interactions with the other protons that are already present in the nucleus. As such, although free neutrons are unstable, bound neutrons are not necessarily so. The same reasoning explains why protons, which are stable in empty space, may transform into neutrons when bound inside of a nucleus."
In some senses, protons exert an "electrostatic pressure" on each other. The bound neutron cannot become a proton, because of this electrostatic pressure applied by its companion protons - if it were to become a proton, it would need to get over the hurdle of that "pressure". In empty space, this "pressure" doesn't exist, so the neutron can easily slide into a proton-electron-antineutrino state.
As an analogy, when you're hot (say temperature t1) you take your jacket off. However, if you are in a really cramped room with a lot of other people, you'll need to become a lot warmer (say temperature t2) to go through the hassle of removing your jacket and elbowing your neighbour in the eye. If the room is air-conditioned so that you only ever get to temperature t1.5 (where t1 < t1.5 < t2), you'll not take your jacket off, even though you would if you were in a free state.
The basic idea is that a neutron all by itself has a lower energy state available to it, which is to decay into a proton, an electron, and an antineutrino. On the other hand, in the nucleus of a stable atom, a neutron decaying would result in a higher energy state for the whole nucleus.
The difference can be explained by potential energy. A neutron decaying in empty space causes a proton and an electron to wink into existence on level ground. A neutron decaying near a proton causes a proton to wink into existence on top of a cliff (the proton's electromagnetic field) and an electron to wink into existence at the bottom of a trench (ditto).
As far as what critical line is crossed, I can't really say. In fact, I'm not even sure if QCD / Electroweak Theory are yet able to calculate that.
