Electronic – Batteries and running time

Does the circuit make sense and do you see it working?

The circuit is rubbish (I'm not noted for my human skills BTW) because the top NPN will only ever produce about 8.2 volts at the emitter to charge the battery - you need to consider using a PNP transistor wired differently.

Also the two voltage sources are floating i.e. their negative connections are floating and please don't whine saying that I should presume they are connected to ground.

Accuracy is something that needs to be a commitment in electronics.

Also, the LED shown in red isn't described as being a 12V type and therefore it might need a current limiter BUT why is it there at all when you have the other LED in parallel?

I also read that you are using lamps but, please get used to the fact that a circuit diagram (and what it contains) is the gospel and, contradicting it with statements about lamps is really just a sign of laziness when using the circuit diagram editor. Get it right in the editor.

In other words, draw an accurate circuit to the best of your ability and let that speak for itself. Transistors all round are probably the best option when it comes to Q3.

Yes, lithium batteries undergo unwanted chemical reactions when discharged below 3V, causing their internal resistance to be permanently and significantly raised. Their capacity will suffer as well, meaning that they won't accept the same amount of charge anymore. When such an overdischarged cell is "brought back to life", it will likely become chemically unstable, creating a risk of a short circuit developing inside the cell.

Even worse, assuming that you measured 2.5V at no load, your batteries have dropped even lower when they were being discharged and have subsequently rebounded to 2.5 V after the load was removed.

Li-ion cells have a maximum voltage of 4.2 V or less, I am not sure where you got the 4.7 V figure from but it's a recipe for fireworks. OP has since edited the question, to a still incorrect 3.7 V. 3.7 V is the nominal voltage (average voltage during a complete constant current discharge), while 4.2 V is the maximum voltage. These figures will vary slightly from cell to cell.

I would completely discharge the cells and get rid of them, 2Ah 18650s are cheap and not worth the risk of them blowing up.