Electronic – Is this an opamp voltage regulator

"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.

1. R30 limits the charging peak current to the capacitor somewhat, but at 1 ohm it still allows for 12 A, so of little use there. Also would limit the current through the zener if there are peaks above 20 V.

2. The larger capacitors work less well at higher frequencies, and that's where the smaller ones take over.

3. 470 µF in a non-polarized version would be expensive, but there would be nothing against it. All large capacitors are polarized.

4. On the output it would give an extra load for the 7805 to charge it. On the input too for the battery, but that can deliver more than enough current.

5. No, except that they're larger.

6. Keep in mind that the input-output difference for the 7805 is about 5 V (12 V - 1 V for the diode and 1 V for R30 - 5 V out) and that at 1 A out the regulator will dissipate 5 W, so it will need considerable cooling (sizable heatsink) if you want to draw that much current.