If the net charges are limited to 1 megaCoulomb, there will be no effects on either planet. The ratio of electrostatic force to gravitational force is
$$\frac{F_e}{F_g}=\frac{kQ_1Q_2}{GM_1M_2}\approx4\times10^{-18}$$
meaning that the electric attraction will be almost 18 orders of magnitude lower than the gravitational force. You can also run the numbers on each individual planet and find, similarly, that it's not possible for a charge of 1 megaCoulomb to tear a planet apart; the gravitational binding energy is just way too high.

For the two to even be within two orders of magnitude (i.e. $F_e/F_g>10^{-2}$), each planet would have to hold a charge of $\sim10^{14}$ Coulombs. That would require a charge imbalance of either $\sim10^{33}$ extra protons or electrons on each planet, which is still tiny compared to the mass of a planet. This seems quite unlikely to happen - I don't know of any mechanism that could strip away that much charge. I'd assume it would require ionizing atoms in the atmosphere and then allowing electrons to escape hydrodynamically or through some sort of handwavy magnetic fields, but freeing bound electrons at the requisite level seems difficult. (I'm also tempted to say "something something stellar winds", but that's really no better an explanation for ionization.$^{\dagger}$)

Were that to happen, I suspect you'd see a difference once the two atmospheres were able to touch, as suggested by PcMan. If much of the extra charge was distributed in the upper atmosphere, and the two planets were approaching each other at a somewhat realistic speed of $\sim20$ km/s, then there would be a period on the order of a couple of seconds between the atmospheres beginning to merge and the surfaces colliding. In that brief timeframe, it's quite likely that you'd see lightning and similar extreme electrical effects in the atmosphere.

At this point, assuming that the $\sim10^{33}$ extra protons on one planet combine with the $\sim10^{33}$ extra electrons on the other, there would be intense emission from recombination. If each electron were to immediately fall to the ground state, we'd see
$$10^{33}\times13.6\;\text{eV}\approx10^{15}\;\text{Joules}$$
which [Wolfram Alpha](https://www.wolframalpha.com/input/?i=10%5E33+*+13.6+eV) says is about five times the energy released by the impact forming Meteor Crater in Arizona.$^{\ddagger}$

This is significant in everyday life, but insignificant compared to the energy released by two colliding planets. You'd also see a significant amount of energy coming from the electrostatic potential energy of each planet - but again, this wouldn't be comparable to the energy released by the collision. Remember, of course, that this requires some absurd process to create the charge imbalance in the first place.

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$^{\dagger}$ That's a slight lie - I can't rule out a process involving stellar winds. I also can't rule out something using ultraviolet ionizing radiation from the parent star, which is important for planets orbiting M dwarfs. On the other hand, this radiation typically doesn't result in such dramatic charge imbalances!

$^{\ddagger}$ In reality, this *won't* happen instantaneously - or even directly to the ground state - but instead quite slowly. Recombination to the ground state would release a photon of energy 13.6 eV, which could go on to ionize another atom, so recombination to excited states would be the relevant processes, just as in H II regions. As the recombination timescale is much longer than the ~1 second between when the atmospheres touch and the planets collide, the atmospheres would quickly be heated up, which in turn would lower the recombination coefficients, which usually go as $T^{-3/4}$. So we'd see a burst of recombination, followed by a sharp decrease, and depending on what happens to the atmosphere following the collision, it could still take some time for the majority of the electrons and protons to recombine.