It sure can. Realize that an LED is going to require at least 10mA but powering it from +12 to -12 means 24V * 0.01A = 0.24W. If you only used one half of the supply, that would be 12V * 0.01A = 0.12A. But Eurosynth usually has +5V I believe in which case you could use that instead and now power is 5V * 0.01A = 0.05W.
Technically you could do even better than this because you could, in theory, make a DCDC converter that converted whatever voltage into just enough voltage required to power the LED which has a voltage drop between 2V and 4V. But just using 5V or a DCDC converter to make 5V from 12V (which is a very cheap part an a tiny package that makes 10's of mA) would be 0.05W which, again, is 5 times more efficient than your circuit.
Now, if you want the LED-on threshold to be near 0V, then you could use a PNP with a large emitter resistor from +12V to -12V and then drive the base of an NPN with LED and Rmax resistor from +12 to 0V. The PNP serves three purposes here:
1) The diode drop of the PNP shifts the NPN bias to compensate for it's diode drop (meaning the LED will start to turn on when the control voltage is just above 0V and not at 0.6V).
2) The PNP stage would consume very little power with it's large emitter resistor (47K would be 24V * 0.0005A = 0.012W) and the NPN LED driver would only consume power between +12 and 0V which would double the efficiency of your circuit with only one additional transistor and resistor. If 5V is available, again, much less still.
3) If Rmax is the NPN emitter resistor, it will provide feedback to yield a fairly linear voltage to current converter. This is good because LED brightness is proportional to current. With just the NPN, you will get an exponential converter which is almost certainly not what you want.
If you restated your objectives more clearly, I think you would get much better answers in this case. You would probably also benefit greatly from using LTSpice. This is precisely the sort of circuit that LTSpice will help you understand.
