Electronic – Opamp operation with “unbalanced” power supplies

"If I assume an ideal opamp, + and - inputs are equal."

That's not automatically true! Equal inputs are not a property of the ideal opamp! You can have different input voltages even with feedback, think the positive feedback of the Schmitt-trigger. You have to construct the right negative feedback loop to get the inputs equal.

The diagram would probably have been easier to understand if they would have drawn the pass transistor's symbol, instead of that rectangle. Like Olin says, that transistor will be a PNP. \$I_{GND}\$ is the PNP's base current, and the collector current to output is proportional to that. So increasing \$I_{GND}\$ won't draw that away from the output, but instead increase the output current.

What you describe would happen if the pass transistor were an NPN. In that case \$I_{GND}\$ would indeed be taken away from the pass transistor's base current, and an increase in \$I_{GND}\$ would result in a decreased base current, and therefore decreased emitter current to the output. In that case the inputs of the error amp have to be switched.

(Add a resistor to the error amp's output to reduce the amp+transistor's transconductance. Without it, even with real-world components, a 1nV input change may result in a 1A output change, and it may oscillate. (Murphy: your amplifier will only oscillate if you haven't foreseen it.))

edit (shunt regulator)
R4 is an important component. If the input voltage is constant then regulating the output voltage means keeping the current through R4 constant. If the load decreases the transistor will draw more current to keep the total current constant. That's not efficient. Load regulation: if the load current would rise the drop across R4 will increase, and the voltage on the non-inverting input of the opamp will decrease. Then the transistor will conduct less to balance the risen load current.

This seems to be a problem I've seen a few times on stack exchange.

Consider an op-amp with localized negative feedback - The manufacturer designs the op-amp so that under the very worst case situations it is stable. The worst case situation is unity gain - this has the biggest chance of being unstable. Anyway, each year the boundaries get pushed a bit more and op-amps improve BUT, why should TI or AD or LT design an op-amp that would be stable with more open loop gain than what the basic device provides? That would be silly (and a marketing/sales disaster) but you (the OP) have created more open-loop gain by inserting Q4 (common emitter) into the output of the op-amp.

Q4's gain will be massive - emitter is grounded therefore the output at its collector will be possibly a hundred times more amplification than what the op-amp produces. Here's what the TL072 op-amp's gain and phase margins look like: -

The red circle is the unity gain of the op-amp and it can be seen that the phase margin (the number of degrees negative feedback is from morphing to pure positive feedback i.e. becoming an oscillator) is about 75º. This is a decent margin but, if gain were increased 30x (by introducing a transistor like Q4), the unity gain point is exactly at 180º i.e. the circuit becomes an oscillator.

Solution - get rid of Q4 and use a rail-to-rail op-amp (or power the op-amp from a slightly higher supply) and restore the feedback to the inverting input. You may ask why the output Darlington-pair doesn't create the same problem - it is an emitter follower with slightly less-than-unity gain and no internal miller effects to shift the phase this way or that.

If you choose an op-amp that can deliver 30mA into the Darlington-pair input then you should be able to get up to 20A from the power supply without the need for Q4.