Led
When a MOSFET is being used a switch, you can connect it in series to the LED across the positive end OR the negative end (in the same way it doesn't matter which side the switch is on for a light bulb. The only catch is that a MOSFET has certain voltage requirements to operate correctly.
ORIGINALLY:
+V-----LED-----(D)NMOS(S)---GND
WHAT YOU WANT:
+V-----(D)NMOS(S)-----LED---GND
(only source/drain terminals of MOSFET shown, the gate terminal on the transistor is now shown)
Unlike a normal mechanical switch where it will operate correctly no matter how you stick it into the circuit, the MOSFET to operate the way you want it to, it has certain voltage requirements.
1. Drain-source voltage has to be maintained above a certain level. You have to examine the circuit and make sure this is true. But this should not be a problem since (assuming it was working in the original circuit, the voltages across the source-drain are the same regardless of the order the components are flipped in).
2. The much bigger problem: The gate voltage turns the MOSFET on and off. The MOSFET turns on when the gate-source voltage exceeds a threshold. Notice, I said the gate-source voltage and not the gate voltage. The gate-source voltage was referenced to GND in the original circuit because the source was connected to ground. So it the gate was at 5V, the gate-source voltage would be at 5V. But the way you want it, the source voltage is no longer at ground, it might be at say, 1V. This means that if you apply 5V to the gate, the gate-source voltage will now be 5V-1V = 4V, which may not be enough to turn the MOSFET on.
Both points 1 and 2 may not be of consequence in your particular situation. If the voltage drop across the LED just happens to be very small, the source voltage might be close enough to GND to allow you to drive the gate through normal means (a driver referenced to GND, instead of a driver referenced to a floating voltage...it get's complicated here).
NOTE: THis may or may not work directly in the circuit on the website you are referring to, since I am not sure how the circuit is built.