When explaining how a car reaches its terminal velocity, is it solely down to how aerodynamic the car is? My thoughts are that if you have a fixed mass $m$, then the contact frictional force will be constant,$ F=\mu{R}$, and because the thrust due to the engine will be much greater than this value, then surely only air resistance will increase the total value of drag? And this is why as the speed increase the acceleration begins to tend to 0 as Drag ∝$v^2$.
[Physics] Terminal Velocity of a Car
aerodynamicsdragfrictionnewtonian-mechanics
Related Solutions
The problem you've formulated is that these two cars are identical aside from the mass difference, so let's just limit this to two identical cars where one has an added weight in it. The heavier car will accelerate slower, based on simple $F=ma$, where $F$ is the same, so $a$ must be smaller for a larger $m$. Friction, which determines the max speed along with the force, is a little bit messier.
The air resistance is basically only affected by the shape of the car, so it will be completely unchanged. The heavier car, however, will have greater ground friction from the contact between the wheels and road and because of that will have a slower max speed. One way you can convince yourself that the tire friction depends on the weight of the car is to just consider that the the tire deforms and creates heat, and greater mass will cause it to deform more every turn. For super fast cars tire friction is actually extremely significant.
Update
Due to increased formalism by the question, I can offer a little more detail. Here is a really basic force diagram I made for this. It's not perfect, but I think it's sufficient. And by the logic of the question, $F_{drag}$ is broken up into 2 parts where one is from the body and one is from the wing. Furthermore, the wing aerodynamic forces are really the wind specific drag plus $F_{wing}$.
So, we were talking about increasing the weight of the car. That increases $F_g$ and $F_{ground}$ because in real life the wheel friction force has a significant dependence on the weight. The question mentioned the need for the wing in order to maintain traction. So, traction is a tricky point, because it is related to the weight.
$$F_g=M g$$ $$traction = \frac{M g+F_{wing}}{M g}$$
It is my belief that if a given turning radius without slipping was needed for a race car, this quantity is what would need to be kept constant. I think this gets to a little of what the question wanted, which is that if the mass, $M$, changed, some redesign would be necessary to maintain the same traction (see the above equation).
At low speeds (more precisely low Reynold's numbers) where the flow is laminar or only partly turbulent the drag varies as $v^\alpha$ where $1 \le \alpha \le 2$. Under most conditions the air flow is turbulent and you can assume a $v^2$ dependance. This will certainly be the case for the terminal velocity of a tennis ball. The Reynolds number for a sphere is:
$$ R_e = \frac{vd\rho}{\eta} $$
The density of air is about 1.2kg/m$^{-3}$ and the viscosity about 1.8 $\times$ 10$^{-5}$Pa.s so for a tennis ball 10cm diameter a Reynolds number of 1000 corresponds to 0.17m/s.
The drag equation is phenomenological rather than derived from any rigorous theoretical treatment, and the drag coefficient is basically a fudge factor. It's only approximately constant. For example the drag coefficient of a sphere can vary depending on the speed and can have values greater than 1 at low Reynolds numbers.
Best Answer
It is just a simplification to say that total drag is $\propto v^2$ and to ignore all other forms of friction. This applies because the power to overcome aerodynamic drag is $\gg$ power loss on the drivetrain and rolling resistance. If you included all the other terms the final speed will decrease only very slightly.
So a simplified model of acceleration (as a function of speed $v$) is
$$ a = \frac{P(v)}{m v} - \beta v^2 $$
Where $P(v)$ is the engine power at speed $v$, $m$ is the mass and $\beta$ is some drag constant.
Top speed occurs when $a=0$ and hence $$v_{top} \approx \sqrt[3]{ \frac{P_{max}}{m \beta} } $$
This applies only the the car is drag limits (and not gear limited) and that it is designed with getting peak power at the rpm corresponding to top speed. Some cars with 6 speeds, may only achieve top speed in 5th gear, and 6th gear is just an "overdrive".