It is a long question, but better than a short one, as you've shown your own research.
1) Solar cells. If you're stacking your own ones, stack 9 of them and get the 4.5V of the original circuit.
2) Battery charging. Batteries are the only thing you've left out of your spec. This is an area where the circuit design relies on cutting a lot of corners. In theory it might be out of spec, if you were to put 4.5V at 280ma through AA NiMH cells indefinitely. In practice, you don't get full sun all day, you'll be using it indoors, and you're not going to get optimal power transfer from the cells, so this isn't going to cause problems.
3) Diode. It's just a regular diode, not a zener. Current through it is actually determined by the battery and right hand side circuit, not the solar panel - the transistor is off when the panel is generating electricity. The original 1N914 will be fine. 1N4004 will also be fine.
4) Resistors: not a precision component here, use whatever meets your cost constraint. 5.1k for 5k is fine.
5) Wire: not critical. Your ebay link looks suitable. Thinner is better for the toroid.
6) Transistors: stick with the exact part numbers. Design may rely on specific parameters.
7) LED: again, this circuit relies on cheating. Normally a white LED won't run from two NiMH cells. The joule thief part provides a boost converter that gives small pulses of higher voltage. It doesn't have the capacity to provide a lot of current at that voltage. In combination with the pulsing this means there should be no risk of damaging it.
(A proper analysis of this circuit would be good, if nobody else supplies one I'll do it in a few days).
I would use a LM555; it's a very flexible, cheap, and easy-to-find timer chip. And you can find lots of examples on how to use it. Here's one to start with. The 555 will source around 125mA, which isn't quite enough for your application. If want to drive all of the strings at once, you can do it with an NPN transitor switch hooked to the output of the 555. I would probably start with a TIP120; that's hefty enough to handle the current you need to switch. You will need a way to power all of the strings at once. If you are using batteries, I would use 3 D cells to drive all of the strings. If you are okay with a plug-in solution, you can find lots of small wall wart power supplies that put out 5V at 1Amp; they are used to charge cell phones and other kinds of electronigs
Best Answer
You MUST specify your battery "chemistry" as "2.4V" could mean several things and results will vary. That sounds like 2 x AAA NimH cells. Are you limited to that capacity or the size chosen or ...?
I'll assume 2.2V = 2 x 1.1V = 2 x average output voltage of a NimH cell at moderate load.
If using NimH then voltage will fall to 2V at end of battery life so system must run off 2V.
At say 2.2V x 1000 mAh = 2200 mWh battery energy and say 80% converter efficiency, for 4 hours operation you can get 2200 mWh x 80% /4 hours = 450 mW.
This is so close to 1/2 Watt as to not matter. You can get 80% "easily enough" - 90% may be possible but is harder across the whole voltage range.
A very common and low cost IC is the LM319 / LM339 dual/quad comparator. These are almost certainly easily obtained where you are.
These will operate with Vcc from 2V to 36V and can be used to make oscillators and to act as switching elements in switching reqgulators. They do not have enough current drive to drive the LED directly but will drive a suitable MOSFET or bipolar transistor plus inductor in a boost converter circuit.
LM319 dual comparator data sheet
LM339 quad comparator data sheet
The circuit below, which is Fig 3 on this page could have the 555 oscillator replaced with an LM339 oscillator and a current sense resistor between LED cathode and ground would provide LED-current feedback control. Components could be adapted to suit.