Great work! This is a very minimal circuit. They did a good job on keeping the cost down. Combining the circuit, and my knowledge of how these little lights work, I will provide this answer which is not totally complete. Basically, I think you are on the right track.
At night, the solar cell is dead, so the current through R1, R2 is very low/zero. D1 is reverse biased, and no current flows through it. T2 is off. The rest of the circuitry is kind of a self-exciting oscillator which is arranged so that it dumps energy into the LED during one polarity of the oscillation.
So now assume it is night time. T2 is off. D1 is not conducting current. Just ignore them. If you start with T1 off, you can see that current would ramp up in the inductors (assume at this stage that the LED current is zero... it takes more than a 1.2V battery to get the white LED to turn on). Voltage at C1 will climb until T1 turns on. Then the current in L1 will ramp up even faster, but the current in L2 will decrease and reverse until the voltage at C1 is too low to keep T1 turned on. So T1 will turn off, and the energy stored in L1 will be dumped into the LED. Meanwhile, the voltage at C1 is climbing again, or eventually will climb when L2 is turned around, which will eventually turn on T1 again, and so-on until the battery dies or something breaks or the sun comes up.
EDIT
Once the sun does come up, D1 will be forward biased, the battery will be recharged by the solar cell, and a small current will flow through R1 and R2. The voltage across R2 will cause T2 to turn on, which will hold T1 continuously off, which will suppress the boost / oscillator circuit.
END OF EDIT
Looking at this, I would not have been able to say for sure whether it would work or predict the average current in the LED. But obviously it does work, so I have just kind of rationalized it after the fact. I like this circuit. It is an interesting way to provide boost for an LED without any fancy circuitry. The BOM cost is dirt cheap. Thanks for posting and interesting question and doing so much leg work!
I picked up a bunch of doorbell transformers (just for my junkbox) at $1 each, a while back. About the same as you describe. I'd probably use a bridge rectifier followed by a ripple capacitor to get the basic DC part done in preparation for an LM2596 buck converter. (You can get these for under a dollar on ebay, if you don't want a fancy voltage and current display, and they can easily handle your LED requirements. Most include potentiometers to use in setting the voltage and current limit, too.) Once you get the voltage into the right vicinity (say 5 V), you can then drop the rest of the voltage with a current limiting resistor in the usual way.
The first thing to do is to test your transformer under load. Just to make sure it works, mostly. I use one of those big 25 W resistors -- \$100\:\Omega\$ if you have one around? Load it down and see what it does. Chances are, it's fine though. But I like to know what the transformer does under load just to be sure.
But that's what I'd probably do (partly because I have a small box of buck converter boards laying around, too.)
Best Answer
There are two pieces that work together in order to light the LED.
The first is an IC such as the QX5251. This is a solar battery charger and LED driver in a TO-94 package. It rectifies the solar input as well as detects when it is dark in order to activate the LED.
The second is an inductor which the driver uses to boost the voltage from the battery high enough to light the LED. The driver uses the same mechanism as a boost regulator but without any of the filtering.
The result is an LED that is imperceptibly flickering. This flickering may interfere with your vision attempts, so do plenty of testing before deploying it in the wild.