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"The electrons play no significant role in nuclear (hence the term nuclear) fission. It is the splitting of the nucleus that matters, and all the resulting fragments are going to have positive charge."

No, and I never said they did. But as soon as those two positively charged nuclei form, they start grabbing electrons from whereever they can get them, including the ones floating around the individual atom. So, probably by the time they have reached their "first bounce", they may be either positively or negatively charged.

But all that aside--the electrostatic repulstion between even two all-positively charged is TINY compared to the energy imparted due to the mass difference. We're talking a few eletron volts vs MILLIONS of electron volts.

I am going to quote a competent nuclear physicist from one of his books. I believe he also had a PhD, but he didn't go around flaunting it.

... The nuclear forces act mainly between each proton (or neutron) and its nearest neighbor, while the electrical forces act over larger distances, giving a repulsion between each proton and all the others in the nucleus. The more protons in a nucleus, the stronger is the electrical repulsion, until, as in the case of uranium, the balance is so delicate that the nucleus is almost ready to fly apart from the repulsive electrical force. If such a nucleus is just "tapped" lightly (as can be done be sending in a slow neutron), it breaks into two pieces, each with positive charge, and these pieces fly apart by electrical repulsion. The energy which is liberated is the energy of the atomic bomb. This energy is usually called "nuclear" energy, but is really "electrical" energy released when electrical forces have overcome the attractive nuclear forces.

So, probably by the time they have reached their "first bounce", they may be either positively or negatively charged.

How would they become negatively charged?

But all that aside--the electrostatic repulstion between even two all-positively charged is TINY compared to the energy imparted due to the mass difference. We're talking a few eletron volts vs MILLIONS of electron volts.

That doesn't sound right. Let's say the protons and mass are split in two equally so that the calculation is easy. Each nuclei would have a charge of about 73.6E-19 C. Let's also say that the two nuclei are separated initially by 10E-15 m. The energy in that electric field would be about 5E-11 J which is about 3E8 eV or 300 MeV. According to this, "a typical energy released by a fission event is approximately 200 MeV" so the electric repulsion of the two main products could easily explain the bulk of it.

However it is clear that not all of the energy released by fission can be explained that way since some of the products (neutrons and photons) are uncharged also carry away a lot of the energy. In fact, IIRC, the neutrons need to be slowed down by a moderator to captured and induce further fission reactions.