In the Norton equivalent example, why did shorting $A$ and $B$ result in the resistance $R_1$ being ignored?

Shorting $A$ and $B$ in this example would result in the two sides of $R_1$ being at the same potential. So there would be no current through it according to Ohm's Law, and so it is irrelevant to the circuit. Alternatively, you can think the shorted circuit as having $R_1$ in parallel with a zero-resistance wire. All the current is going to go through the wire and nothing is going to go through $R_1$.

Note that you can't ignore the branch with the voltage source and $R_2$ when $A$ and $B$ are shorted. The reason is that the voltage source will be pumping charge to maintain a fixed voltage across it. To keep the two ends of the branch at the same potential, there must then be a voltage across the resistor $R_2$ and hence some current flowing through that branch.