Why does current flow into the transistor in a current shunt monitor?
Let's assume that the op-amp has very high input impedance and that no current flows into the + and - pins. If current is flowing through Rs then the + input voltage will be higher than the - and the op-amp output will start to go high. This will turn on the output transistor allowing current to flow through RG1, the transistor and RL. This will cause a voltage drop across RG1 and the circuit will settle when the voltage on + input is the same as that on the - input.

Since it's a real op-amp the input impedance is not infinite and some current will be drawn. By making RG1 and RG2 the same the input currents and voltage drops will cancel out.