Im speaking directly to the initial basis of the argument; the force of friction (only thing stopping the car) is 100% F = coefficient of friction x mg. No argument.

but the element that the coefficient of friction of an air filled tire is independent and inversely proportional to the mass/load on said tire, changes this entire equation. The reason the cars would stop at the same distance, from our original discussion, was because the friction force is different due to the normal force being less on a lighter car. And assuming the coefficient of friction (originally attributed to tire compound alone) was the only constant, this would mean that mass cancels out of the kinetics equation; the two cars stop in the same time and distance.

But if the coefficient of friction loss due to mass/load on a tire negates the mass effect on the normal force, suddenly the Force of friction IS constant, and mass remains in the kinetics equation.

F=ma

F=force friction

F friction = coeff x mg

coeff = compound coeffecient / masss of object loading tire (again... I don’t know if this is 1:1 and if you know it isn’t or it’s scaled this would change)

F friction = (compound tire coeff / mass of object) x mg

Force of friction in this instance then being only compound of tire x gravity

compound tire x gravity = mass x acceleration...

mass still would be a factor in stopping distance