The pressure you are referring to is the internal pressure of the tire. The external pressure is very different since the surface is not smooth and the tire contains the internal pressure. For example, when the car is up on jacks, the internal pressure is the same but the ground pressure is zero.
From a physicist, I find that a puzzling answer.
The pressure inside the tyre is constrained by tension in the carcass of the tyre. If the tyre is on the ground with a load on it (the car), the tyre deforms until a state of equilibrium is reached. This point is when the force acting from the ground upon the tyre equals the force required the deform the tyre.
Clearly, if there was no pressure in the tyre, the only force resisting collapse is that due to the strength of the sidewall. As the tyre is flexible, this is not enough. The additional force due to the air pressure inside the tyre stops the rim touching the ground.
Thus force exerted upon ground = pressure within tyre + force required to deform sidewall.
If the tyre was rigid, then the strength in the sidewall alone would be able to withstand loads without deforming. But it ain't, so it doesn't.
As a side point, it is interesting to find (well it interested me, anyway) that the pressure inside the tyre stays pretty much the same regardless of whether there is a single person in the car, or 5 people plus luggage. This is because the contained volume within the tyre stays pretty much the same. What does change is the size of the contact patch.
I once carried out the following experiment:
Put a piece of graph paper on a set of bathroom scales. Jack the car up, and coat part of the under side of the tyre tread with some printing ink. Lower the tyre onto the scales and record the reading. Remove graph paper and count the squares within the contact patch. Repeat for various loads and pressures.
Within the limits of the experiment (I didn't test to the point of destruction), the results will show that the size of the contact patch changes with the load, and in inversely with the pressure. The pressure inside the tyre hardly changes at all in relation to the load. Conclusion: the overall load on the ground increases as the load increases, but as the tyre deforms, the contact patch increases in size, and the force per unit area stays the same.
I think you assumption about very near the melting point is reasonable.
Don't know much about skates, but if the blade has a contact area of, say, 1/8" x 12", then a pair of skates supporting someone of 120 lbs will be exerting a force of 40 psi on the ice. From the above, this would seem similar to the unit force of a typical tyre upon the ground. But I imagine there would be other factors invloved in liquifying the ice, like its temperature, for example.