The problems started when the USB connector of the original device got damaged. Would it be reasonable to repairing the connector? You could also try to bypass the original connector to feed the charger.

Separate charger approach

The proposed approach with separate charger is sound. However, I would look for a charger, with:

• A bit less current. Lithium-ion cells should be charged at current of 0.2C to 1.0C. This means that capacity of the cell drives the charge current and charge time. 0.2C is a shorthand which means that the current in [Amperes] is 0.2 of capacity in [Ampere-hours]. Units don't quite match, but again, this is a shorthand. 1A is on the upper side of the range. I would look for something between 0.2A and 0.6A.
• Published schematic or known part number for the charge controller IC. This improves the chances during troubleshooting.

At the same time, there's nothing fundamentally wrong with the charger, which you've found on eBay.

The battery pack connector looks like Molex PicoBlade. Here's a 4-pin male (p/n 53047-0410).

Yes, you can charge the two cells as a series 2.4V pack. During charging the voltage will rise to ~3V.

The charger should be current limited, to avoid charging the battery too fast and overheating it. The circuit could simply be a resistor in series which limits current to a safe 'trickle' level (10 hour rate = 100mA for a 1000mAh battery). The resistor might be inside the charger (perhaps explaining why the voltage drops when you try to draw power from it) or built into the iron - then the 'charger' is just a DC power supply with high enough voltage to make up for loss in the resistor.

While you have the iron disassembled you could look for a current limiting circuit. If it doesn't have one (ie. the battery is connected directly to the charge socket) then you must use a charger which is current-limited. Don't try to use a DC power supply, even if its ratings appear to be the same.

Most USB ports will deliver 100mA without any negotiation, so to charge from USB you just need a resistor in series which limits current to <=100mA. Assuming the cells are 1.1V when flat, the resistor must drop 5V-2.2V = 2.8V. Calculating the value using Ohm's Law gives 2.8V/0.1A = 28Ω (the nearest preferred value of 27Ω should be close enough). The resistor will dissipate up to 2.8V*0.1A = 0.28 Watts, so it should be rated at 0.5W or higher. As the battery charges up the voltage difference will reduce causing charge current to drop, so to get a full charge you may have to leave it on for 12~14 hours.

The original Nicad battery probably had lower internal resistance and could hold a slightly higher voltage under load, so you may find the iron takes a bit longer to heat up. However NiMH has higher capacity than Nicad per volume, so you might considering using similar size NiMH cells that have higher capacity - rather than smaller cells which have higher internal resistance and may not last as long. Avoid ultra-high capacity AA cells, as these are optimized for capacity rather than power.