Either method works, but I would generally use the first because it is simpler and requires fewer parts.
If the signal is actively driven both directions, like from a CMOS digital output, then you don't need the resistor between base and ground. In other words, you can lose R2 in the first circuit, and R10, R11, R12, and R13 in the second.
As you say, the transistor in the first circuit needs to be able to handle the combined LED current. For four normal indicator LEDs, that is not much of a limitation.
Depending on the voltage of the LEDs, you might be able to drive two at a time in series with a single resistor. For example, if these are red with 1.8 V forward drop, and the transistor goes to 200 mV in saturation, then there is still 1.2 V left for a resistor to set the current. Doing that gets you the same LEDs lit with the same brightness, but with half the current used.
Green LEDs with 2.1 V drop are on the edge, but can work doubled up. Two LEDs would drop 4.2 V. Again, figuring 200 mV for the saturated transistor, that still leaves 600 mV for the current limiting resistor. Especially if you're not trying to run the LEDs at their limit, this can be a legitimate current savings.