An ideal bulb is not infinite or zero resistance.
An ideal bulb is an ideal resistor. No parasitic capacitance or inductance, just resistance and no change with temperature.
So with all ideal parts, and assuming nothing near the wires to absorb near field radiation, the resistance of the bulb is all that absorbs energy. So it will eventually settle, still an asymptotic delay. I think you could model this as two paralleled transmission lines shorted at the far end, so simplify it to one shorted transmission line of half the characteristic impedance with the battery and bulb at one end.
So the wave has to travel to the other end and back again, say 5 times.
However, if we arbitrarily make the bulb resistance equal to the characteristic impedance of the paralleled transmission line, then it should immediately (1m/c) reach 1/2 power. And reach full current/power in the time it takes for the step change to propagate to and back from the shorted end.
Since it is all perfect parts, a shorted transmission line reflects back in opposite polarity, therefore the reflected step change is in exact peak value but opposite polarity. So as it propagates back to the bulb, it exactly cancels the voltage present.
When it reaches the bulb/battery end, it is zero. So full brightness after length of wire/c. So if the wires are 1 light second long, it takes 1 second to reach the end, 1 second for the reflection to return. So the bulb will reach 1/2 brightness in the time set by the spacing of the wires, and 2 seconds later it steps to full brightness.