Well, they would have full voltage across them whenever they turn off, I think, but when on, they (like phase FETs in controller) would only have the voltage created by the current flowing thru it's RDSon. So if they were say, 20milliohm FETs, and current was 2A, it'd only be 40mV across the charging FET (since it's just one, no parallelling). That's only 80mW, so there's no reason for it to get hot at all, during charging.
What might be happening is one of these things:
1--it is not melting during charge, but instead is coming off during discharge when *those* FETs heat up (though again, they shouldn't get that hot, based simply on Ohm's Law, unless they are buried inside the shrinkwrap with no airflow at all, heat trapped right there). If the other FETs got hot enough, they could heat it up enough to come off with them even though it's not in use.
2--Again, buried in teh insulation, perhaps that 80mW (maybe 100-200mW worst case with a beefier charger?) might be enough over a charge cycle ot heat it up that much.
3--The heat build up causes RDSon to rise, creating more voltage drop and thus more waste heat wattage, inside the insulated pack that then causes it to get hot enough to unsolder or fail
4--If a voltage spike during charge switchoff (HVC) kills the FET, or damages it causing even higher RDSon than usual, then #2 and 3 above would be even more likely.
Or something else I haven't thought of.