Why is the neutral tied to ground in USA single phase wiring?

If you're drawing real power, then you're creating a load that is at least partially resistive - this shows up as a slight phase shift in the relation between voltage and current, both in the primary and the secondary - which results in a slightly resistive-looking load. If you don't, the P=IE formula integrates to have no real component, and thus no real power.

All the hand-waving in the world won't change the laws of physics. You can mess with the imaginary power all you want[*] (inductive and/or capacitive) but if you want work done, you need to create some sort of at least partially-resistive load, which is reflected onto the primary, which causes losses.

[*] sort of. If you're non-resistive by enough, they ask you to put compensators at your site to "normalize" the load. Even so-called imaginary power costs money, because the current through the wires causes resistive losses.

Drawing real power from a cable within the field will create its own field which induces a voltage in the transmission line. This opposes the current, requiring work to be done.

There is simply no way that you can reduce either the real or apparent power in the transmission line by extracting real power from it.

Drawing apparent power (i.e. drawing current 90 degrees out of phase with the voltage) can either increase or decrease the apparent power of the transmission line depending upon whether it shifts the phase of the current toward or away from the phase of the voltage.

But you can't power anything with apparent power.

And any such parasitic power draw is *in addition* to the existing transmission losses. Adding additional loss doesn't make any existing loss magically disappear.

It doesn't work like that. An AC transmission line through the middle of deep space won't be losing anything. Creating a magnetic field doesn't (in itself) require any power. Extracting power from that field does.