"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.
In the negative feedback configuration, the op amp will drive its output such that the voltage between its positive and negative inputs is (ideally) zero. In the first example, the positive input is at 3V. Therefore the op amp will drive the negative input to 3V. So R1 will have 2V across it, and therefore 2mA through it; this 2mA all flows through R2 (no current flows into input because an ideal op amp has infinite input impedance), so there is 4V drop across R2. Therefore the output is 3V - 4V = -1V. Apply the same reasoning in the second example.
You can solve for the output voltage symbolically by summing the currents at the negative input:
$$
\frac{V_3-V_4}{R_1} + \frac{V_{out}-V_4}{R_2} = 0
$$
Best Answer
I think the designer (grad student) was having a bad day....
MC3401 has NPN inputs unlike the quad LM324 with PNP inputs that operate to Vee =0V.
The 3401 must have Vin+>Vin- by having a slight +ve input offset current.
Otherwise if Vin->Vin+ the output goes low and stays there no matter how much negative feedback.
When this condition occurs. the Zener current, Iz causes Iz*R1 = __ mV for the stable low Zener current then near null offset and Vout=Vz @ Iz
Since Vin+ has lower input bias current than the collector Vin- then Vin+ will have a lower input voltage than Vin- which forces the output at 0V.
So Vin+ must never be 0V=Gnd. It must be slightly > Vin- otherwise the Norton or any Op Amp will not work. This takes 3 R's/ Pull up/down and series R.