The answer is that the whole circuit is full of electrons. I think you may be thinking along the lines of "if I switch a tap on, the water takes time $v/L$ to reach the end of a hose of length $L$. So, if I switch a light on, the electrons must take analogous time the reach the light". Because the circuit is full of electrons, the energy source shoves the electrons near it, which shove electrons further down the line and so on. The energy propagates as an electromagnetic wave. The electrons which first light the light are not the ones that first go through the closed switch.
If you switch on a tap with "slow water" going through it and the hose is full of water when you begin, the water comes out the end at speed $v$ as soon as you switch the tap on (or it would, if the water were incompressible: in practice an acoustic wave runs up the hose and accelerates all the water, so that there is a tiny delay - but it's much, much less than what you'd reckon from $v/L$).
ANother way to put this: you can shove something quite a distance away from you with a rigid bar and it begins to move straight away, even if you are pushing the bar slowly.
Actually, the analysis of what really happens when you switch a light on is one of the most fascinating exercises in electromagnetic theory. The energy does NOT actually go through the wires at all! There is an explanation for this in the Feynman Lectures on Physics - I can't find it right now. What happens is this: when the switch closes, the electrons rearrange themselves on the wires so as to guide the electromagnetic field. They can do this very swiftly notwithstanding their low drift speeds by the mechanisms I have just described. The electromagnetic field then transports energy through the free space around the wires - if you look at the Poynting vectors in the wires, you'll find that there is zero power flux through them.