"...but the moon does not have any linear momentum due to rotation"

This is not correct. Why?

At the moment a tractive force is applied through the wheels, and an equal and opposite force is applied on the moon. Newton's 3rd law still applies.

This force accelerates the center of mass of the car and the moon. Both acquire linear momentum.

In fact, total linear momentum is conserved here.

When you consider the details, you will see the car will orbit around the common center of mass (blue cross below) while at the same time the planet will also orbit around the center of mass at a smaller distance.

If the car of mass $m$ is orbiting at distance $r$, and the planet of mass $M$ is orbiting at a distance $R$ then you can say that

$$ m\, r = M\, R $$

This combined center of mass remains fixed (is an inertial reference frame) as no external forces are applied to the system.

As the car accelerates going faster around its orbit, the planet must counter accelerate also increasing its orbital speed also. Both objects must remain diametrically opposed with respect to the common center of mass in order to maintain the above $m\,r = M R$ relationship.

"...but the moon does not have any linear momentum due to rotation"

This is not correct. Why?

At the moment a tractive force is applied through the wheels, and an equal and opposite force is applied on the moon. Newton's 3rd law still applies.

This force accelerates the center of mass of the car and the moon. Both acquire linear momentum.

In fact, total linear momentum is conserved here.

When you consider the details, you will see the car will orbit around the common center of mass (blue cross below) while at the same time the planet will also orbit around the center of mass at a smaller distance.

_{Above $R$ is shown as a negative value.}

If the car of mass $m$ is orbiting at distance $r$, and the planet of mass $M$ is orbiting at a distance $R$ then you can say that

$$ m\, r + M\, R = 0$$

This combined center of mass remains fixed (at the origin) and is an internal reference frame as no external forces are applied to the system.

As the car accelerates going faster around its orbit, the planet must counter accelerate also increasing its orbital speed also. Both objects must remain diametrically opposed with respect to the common center of mass in order to maintain the above $m\,r + M R = 0$ relationship.

If at some time the car has velocity $v$, then the planet must have velocity $V$ in the opposite sense such that

$$ m v + M V = 0 $$

which is interpreted as the total momentum remains constant (at zero).

In terms of accelerations, if the acceleration of the car is $a = \frac{F}{m}$ and the acceleration of the planet is $A = \frac{-F}{M}$ then the above is

$$ \begin{gathered} m a + M A = 0 \\ m \frac{F}{m} + M \frac{-F}{M} =0 \\ F - F = 0 \end{gathered}$$

Which is exactly Newton's 3rd law.

You might decide to do a more rigorous analysis based on the following free body diagram

but remember that both objects orbit, and thus both have centripetal acceleration equalling $\frac{v^2}{r}$ and $\frac{V^2}{R}$ respectively. Then you can do the balance of forces in the x and y directions to see for yourself how the system behaves.

"...but the moon does not have any linear momentum due to rotation"

This is not correct. Why?

At the moment a tractive force is applied through the wheels, and an equal and opposite force is applied on the moon. Newton's 3rd law still applies.

This force accelerates the center of mass of the car and the moon. Both acquire linear momentum.

In fact, total linear momentum is conserved here.

"...but the moon does not have any linear momentum due to rotation"

This is not correct. Why?

At the moment a tractive force is applied through the wheels, and an equal and opposite force is applied on the moon. Newton's 3rd law still applies.

This force accelerates the center of mass of the car and the moon. Both acquire linear momentum.

In fact, total linear momentum is conserved here.

When you consider the details, you will see the car will orbit around the common center of mass (blue cross below) while at the same time the planet will also orbit around the center of mass at a smaller distance.

If the car of mass $m$ is orbiting at distance $r$, and the planet of mass $M$ is orbiting at a distance $R$ then you can say that

$$ m\, r = M\, R $$

This combined center of mass remains fixed (is an inertial reference frame) as no external forces are applied to the system.

As the car accelerates going faster around its orbit, the planet must counter accelerate also increasing its orbital speed also. Both objects must remain diametrically opposed with respect to the common center of mass in order to maintain the above $m\,r = M R$ relationship.