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).
The significant difference here his where the currents go.
With a BJT it doesn't just "act like a switch" - that is the end result you see at the high level, not what it actually does.
What actually happens is you apply a current to the base, and that then flows through the transistor and out of the emitter down to ground. At the same time it allows proportionally more current to flow from the collector down through the emitter to ground.
We'll call these two currents \$I_B\$ and \$I_C\$ for the Base and Collector currents respectively.
So in the left hand schematic, the \$I_B\$ flows through the base resistor, then the BJT, and to ground. \$I_C\$ flows through the LED's resistor and the LED, then the BJT, and to ground. \$I_B\$ is set by the base resistor alone.
But, in the second circuit, \$I_B\$ flows through the base resistor, through the BJT, then through the LED's resistor and the LED. \$I_C\$ also flows through the BJT then through the LED's resistor and the LED. So the LED and its resistor get both \$I_B\$ and \$I_C\$.
Consequently the current \$I_B\$ is being set by both the base resistor and the LED's resistor, and also the voltage drop across the LED. Also the voltage drop across the LED's resistor is not just from \$I_C\$ but from both \$I_B\$ and \$I_C\$ combined.
So from an easy to understand point of view the left-hand circuit is best. However, the right-hand circuit does have some advantages. Mainly because the base current \$I_B\$ is set by two resistors it may be possible to remove one of them (the base resistor) and just have the one resistor in the circuit to limit both the base current and the LED's current. That can save on parts, and if you have a lot of these circuits in you design that reduction in parts can soon add up.
Best Answer
LEDS need a series resistance to control the current passed through them. The resistance 'drops' the excess voltage. Without that even a slight over-voltage will cause current to exceed the rating, and destroy the LED. I have disassembled some keychain lights. The better ones have a resistance or even an electronic current control chip, but cheap ones rely on the tiny batteries having enough internal resistance to limit the current to a value safe for the LED. If your light is one of these, the LED will probably be over-run for the first few minutes of operation, until the batteries deteriorate and reduce their available current. Overrunning a LED tends to make the emission spectrum broader, with a slightly longer dominant wavelength.
The LED will emit a band of wavelengths, with the highest output being at 405nm, which is visible violet light. (Most of the inexpensive so-called UV leds are actually visible violet emitters, which have an emission spectrum that may include some near-UV). Both the UV and visible violet will excite fluorescent or phosphorescent materials. Visible violet and near-ultraviolet light have slight tanning effects on skin, and are suggested to be involved in premature ageing of skin. Those wavelengths are also have some germicidal and photochemical effect, so are potentially harmful to unprotected living tissue, like eyes. However, sunlight contains a significant amount of 405nm radiation, so limited or low-intensity exposure is not considered hazardous. This type of LED emits so much visible light that discomfort and blinking is likely to prevent excessive eye exposure at close range. At a metre or more distance the hazard is negligible. Prolonged close exposure might eventually cause injury.