Battery and virtual ground
The problem is that your feedback signal is taken from the wrong place.



Figure 1. Corrected feedback loop. (Red = remove, green = add.)

In your original layout the feedback signal is monitoring the output of the opamp but has no idea what's happening at the load which is where it matters.
If the circuit is working the voltage at RL2 would be the same as that on pin 3. I think that if you measure it you will find that it is out by 0.7, the emitter-base voltage drop of Q2.

Instead, take the feedback from the junction of Q1 and Q2's emitters. The opamp will now correct until that point is at the same potential as pin 3.

Note that this arrangement can be problematic in some situations (push-pull audio amplifiers, for example, because the to switch from sourcing base current for Q1 to sinking base current from Q2 the opamp has to suddenly switch from +0.7 V to -0.7 V. To solve this a biasing arrangement is used to pass some "quiescent current" to keep both transistors slightly on near this point. It may not matter in you application.