MOSFET doesnt have Base Collector Emitter. it has Gate Source Drain.
You would tie the gate into the OE circuit. You would then use a small 1k pullup from Gate to Source. Source tied to 5V. Drain tied to LED anode, LED cathode to ground.
all you need. IF you want to feel safe, you can put a 100ohm resistor between gate and main circuit. But if you use a standard mosfet that has a +/-10v or +/-20v Gate voltage, a resistor is NOT necessary, except a pullup, however the resistor offers some protection in case the MOSFET decides to ever short, it wont hold your OE line, becuase if it shorts, not only would the LED remain lit, but your OE would be forced high through the Gate to Source short, if one occurs. A gate resistor would prevent a hard short, so there is a chance the OE circuit could still pull low.
In a condition without the resistor, and the MOSFET shorts, OE could certainly try to pull low in that situation, but the pulldown transistor would overload and short out. leaving the OE stuck forever because the resistor isnt there to provide ballast from a hard short. So a 100ohm or so gate series resistor would be a good idea in case of a MOSFET short, so it dont blow your OE circuit.
in a 5v circuit tho, its VERY UNLIKELY the mosfet driving the LED could ever short, but given these chinese junk parts of today, you never know.... just design a circuit with that in mind... Dont need a chinese peice of shit taking out a pulldown transistor in a VLSI chip if it ever decides to short

If you opt-out on the resistor to favor more transistor saturation, if it shorts, youll know becuase the mac wont startup and the VLSI will get hot fast.