Why the resistor in this op-amp circuit?
It is a load resistor probably. Often op-amp circuits have the "load" placed as a resistor.

The current an op-amp drives can greatly increase it's non-ideal property. The finite gain becomes more apparent as you drive more current, along with the output resistance.

When simulating an OP-amp you should always attempt to place a load resistor across the output for the effective load you are connecting to. If you want to look up a method of doing something like this, Thevenin equivalent circuits are a good example.

Community Wiki is on if anyone wants to expand.