Q:

How can an engine provide more torque at the point of contact than the ground's response by static friction without slipping?

In response to the top answer here: Newtonian Mechanics - How does something like a car wheel roll?
I clearly fundamentally misunderstand something here. I can't reconcile there being an angular acceleration about the tires (and thus, a net torque) and the torque provided by the engine being equal to the torque in response to static friction. If the torque provided by the engine was greater than the torque in response to static friction, wouldn't the wheel slip?
update: My confusion was resolved in chat with @V.F., and for anyone else who might have had the same simple misunderstanding in this system, it came down to a misunderstanding of contact forces. 
As stated, the response from static friction is equal to the force of the tyre on the road, but the force of the tyre on the road (and of the road of the tyre though static friction) is NOT equal to the force of the engine on the tyre. This is because the tyre delivers the force to the road through contact, meaning it must physically accelerate to bring the atoms closer to the atoms of the tyre on the road, resulting in both the tyre and the road experiencing a force in opposing directions via Coloumbs Law. Therefore, a little bit of the engine force necessarily goes into accelerating the tyre into the road, and the rest of it is is then delivered between the road and the tyre as a contact force, through static friction. This delta is what results in the constraint of rolling motion, as the tyre experiences both a clockwise rotational acceleration from the engine to bring into contact with the road, and then a forward translation via static friction. 
This was easier to understand in the simpler translational case that is typically used to illustrate the difference between contact force and input force. Imagine two boxes, A and B, of equal mass, M, on a frictionless surface. If you exerted a force, F, on box A, it's clear that both boxes would equally accelerate as one system over the surface - but if you exert a force on box A, and box A is in contact with box B, wouldn't box B respond with an equal and opposite force? Then why would box A accelerate at all? i.e. why isn't the force, F, perfectly transmitted through box A to box B? Well, because box A and box B are, of course, not actually in contact. The atoms at their respective surfaces are at some mutual equilibrium caused by Columbus Law. So, to transmit force F to box B would require accelerating box A into box B - i.e. force F is actually distributed between box A and box B based on their respective masses, as "contact" in this sense requires the acceleration of box A into box B, and thus the contact force between box A and box B is necessarily less than the input force F.

A:

We can treat a driving wheel as just another gear that has to pass the torque down the line. 
Ideally, the torque it passes should be equal to the torque it receives, but, since the wheel has some mass and, therefore, some moment of inertia, a little torque differential is needed to speed it up. At steady state, though, the torque on both sides should be the same.
The wheel may slip, if the drive torque exceeds the maximum torque that can be supported by static friction, which can happen, for instance, on a slippery surface (low static friction) or when the car is overloaded or somehow is blocked from moving or there is an attempt to accelerate it too fast. 

