Does gauge pressure or absolute pressure define tire characteristics?

I am not sure, how that link contradicts that. The link says, that you have to change the pressure in your tire according to elevation. Because for the same amount of air in your tyre, the gauge pressure changes according to change in elevation . So, you have to add or remove air so that the gauge pressure remains constant. Hence, it is consistent with your intuition that gauge pressure is what matters.

OK after many good comments and a bunch of edits, here's what I believe. It's a mixture of both.

The gauge pressure will affect the shape of the tire without a load on it. The deformation of the tire once load is applied will depend on both the gauge and absolute pressure. And the distribution of force across the footprint will depend more heavily on the absolute pressure.

Begin by inflating the tire to a given gauge pressure, but without load from the car. The sidewall will attain a certain stiffness which is due in part to the gauge pressure.

Next, lower the car down so that the tires start supporting the weight of the car. Initially, this support will come from the sidewall of the tire, as the wheel presses it down and the ground supports that.

If the sidewall is strong (due to the material) and rigid (due to the gauge pressure) enough to support the car, then it won't deform and we're almost done. All that is left is to analyze the force of the middle of the tread on the ground -- this will be due to absolute pressure.

But if it is not, which is frequently the case, the tire will deform. It will only stop deforming when something has changed such that the sidewall can now support the full weight of the car.

There are two routes to this: the sidewall becomes more rigid, and some other force augments the force the sidewall's internal force applied upward to the rim.

The former will be true if the gauge pressure increases when the tire deforms. This is likely to be true for most tires that have a cross-sectional aspect ratio which is already squashed compared to a circle, since squashing it more will reduce the internal volume (and tires aren't balloons, they generally do not stretch). So this will play a part.

The latter will also be true because the deformation allows for some of the internal air pressure to push upwards on the sidewall surface that is under the rim. This will mean that the downward-facing surface area becomes larger than the upward-facing surface area on the top of the inside of the wheel, so there will be a net upward force from the air pressure. It will also change the stresses on the sidewall so they are no longer purely compression, so I feel like the full solution could be complicated.

So to recap: I believe the full answer to what shape the tire takes is complicated. However, once it has taken that shape, if you want to look at the contact patch and how evenly the pressure on the road is distributed, I believe the absolute pressure plays the largest role in that.