Push-pull digital outputs require no pull-up or pull-down resistor. They are outputs that produce either a high or a low voltage close to the supply voltages.
Open-drain digital inputs are a little bit of a contradiction unless one presumes it is a digitally configurable input/output where the output is an open-drain and not push-pull configuration. If used as an input I suggest a 10k pull-up resistor. If used as an output 10k may do but so might 1k if the output is expected to switch greater than a few hundred micro-amps.
One fundamental problem is that you're probably (you haven't said what your LED current is) not operating the MOSFET within its Safe Operating Area. Never go by the 'headline specs' on page-1 of a datasheet. Check page-4 Fig.3 of the datasheet, for DC operation (curve-3 or 5), at say 15V (i.e. car battery is charging), this MOSFET doesn't want to be carrying more than about 50-60mA. With 100ohm in series with your LED, knowing nothing about the LED, my suspicion is that you're at or over that limit, in which case MOSFET death is just a matter of time.
But you may be right, it may be user installation SNAFUs that's causing MOSFET failure. At the very least I'd add a 'polyfuse' (a PTC self-resettable fuse) between the LED & the MOSFET, to provide some over-current protection. Places like Littlefuse & Belfuse have plenty of guidance on selecting the right thing here.
I'll assume you're using a +5V PIC24 on a %V Vcc, and not a 3.3V one, as that MOSFET won't be saturated on with a 3.3V PIC's GPIO. I have a disturbing feeling that if you're quibbling about a single protection resistor that you may not even have decoupling caps on the MCU's Vcc-to-Gnd pins nor on the input & output of the Vregulator. This is bad at the best of times, but especially in the hostile environment of a car.
Selling a product that customers have to install themselves, without "protection" of various kinds, is a recipe for disaster. Customers always reverse the polarity when connecting things like this, or as you guessed, put 12V into the MOSFET's D, and other atrocities that are an unavoidable challenge to try to mitigate. This is one of the differences between a hobby-project and a product :)
Best Answer
Ignoring the ins and outs of how mosfets work. The output acts like a switch to ground. It cannot produce a high (+5V) signal by itself. The pull up resistor is used so that when the switch is open the output will be high. When the switch is closed the output will be low (0V)