Yes, the vehicle can definitely move opposite to the treadmill direction, but you do not actually need energy storage to achieve this; all you need is sufficient friction between the wheels and the treadmill and the static surface.
I can show the same behaviour with gears as you can just not get rid of elastic deformation or gravitational energy storage (if some parts lift up) in real world but what I can do is reduce the friction at the back wheels so that the back wheels slip before the front and then this is what happens https://odysee.com/@dacustemp:8/stick-slip-removed-from-front-wheels:0
You may need some kind of energy storage to be able to accelerate, but I do not think it is necessary for steady-state operation.
(I'd need to model the entire system, including friction and momentum and losses, to be able to examine the system phase space, and determine whether acceleration is possible without energy storage or not.)
Also: aw crappola, I topsy-turvied myself with the gearing ratio.
I thought inverse to what I wrote. It is a typical error for me, too. Dammit.
Teaches me right for not testing it with a model first. Apologies.
If we ignore losses, it makes no sense to look at the forces involved. The mechanics of the situation suffice.
If we use \$v_t\$ for the speed of the surface of the treadmill (going in the opposite direction), \$v_d\$ for the speed for (the surface of) the driven wheel on the treadmill, and \$v_c\$ for the speed of the car itself (and therefore also (the surface of) the driving wheels), the machine requires
$$\left\lbrace \begin{aligned}
v_d &= v_t + v_c \\
v_c &= \lambda v_d \\
\end{aligned} \right.$$
where \$\lambda\$ is the gearing ratio, i.e. number of turns of the driving wheels per each turn of the driven wheel, assuming their diameters is the same.
The lower equation describes the locked gearbox.
The upper equation describes that the driven wheel must have a surface speed that is the sum of the treadmill surface and "car" speeds, or something has to give (slip).
Combining the two, and solving for \$v_c\$, we have, in the steady state ignoring any losses,
$$v_c = v_t \frac{\lambda}{1 - \lambda}$$
so there is a steady state for \$0 \le \lambda \lt 1\$ only.
In your example, \$\lambda \gt 1\$, and there is no steady state.
However, do check what happens if you reverse the gearing. For example, with \$\lambda = 0.6\$ (driving wheel surface speed 3/5 of that of the driven wheel, or if their diameters are the same, the driving wheel turning 3 times in the time the driven wheel turns 5 times), the solution is \$v_c = 1.5 v_t\$, i.e. the car runs at 1.5 times the treadmill speed.
When \$\lambda = 0.5\$, \$v_c = v_t\$. When \$\lambda = 0.2\$, \$v_c = 0.25 v_t\$.
I do believe that losses can be adequately modeled by reducing the gearing ratio. If the actual ratio is \$\lambda\$, then the model \$\lambda^\prime = (1 - \epsilon)\lambda\$, where \$\epsilon\$ represent the overall losses in the system, should match physical results.
A real world vehicle does probably need energy storage for acceleration and to avoid tiny losses causing a catastrophic failure, and an adjustable gearing \$0 \lt \lambda \lt 1\$, so it can optimize the gear ratio to the current car velocity \$v_c\$. Achieving any ratio \$\lambda \gt 0.5\$ is sufficient to travel faster than the treadmill.
In case anyone is wondering where the necessary energy is coming from, the answer is obvious: from the treadmill. When the "car" is placed on the treadmill, the larger \$\lambda\$ is, the more energy is needed to keep the treadmill moving at a given velocity \$v_t\$.