Assuming both cars are completely identical with only exception of having different mass, then ideally both cars should have the same final velocity and hence travel the same horizontal distance. I say ideally because there might be some other effects, such as change in coefficient of friction with mass, but simple friction models do not account for this effect.
The free-body diagram of the car on the ramp would result in
$$ma = m g \sin\theta - \mu m g \cos\theta = mg (\sin\theta - \mu\cos\theta)$$
where $a$ is acceleration along the ramp, and $\mu$ is coefficient of friction. Note that rotational motion of wheels has been neglected here. As you can see, the mass appears on both sides of the equation and cancels out, i.e. the car acceleration does not depend on its mass
$$a = g (\sin\theta - \mu\cos\theta)$$
Both cars should have the same final velocity at the bottom of the ramp. Once the car gets horizontal, the free-body diagram would result in
$$ma = -\mu m g$$
and the horizontal displacement is
$$\Delta x = -\frac{\mu}{2} g t^2 + v_f t$$
where $v_f$ is velocity the car had at the bottom of the ramp. There could be some mass-related effects at the point where the car transitions from the ramp to the horizontal surface, but it is difficult to comment without specifics.