Your problem is that you have a circuit that must be driven by an output stage with particular properties, notably a very low voltage for a logic 0, because the capacitor must be discharged down to below the Vbe treshold of the transistor. (You probably should have mentioned your reason for this rather unusual LED driving circuit in the original question.)
You can solve your problem in (at least) two ways: change the driving circuit, or change the LED driver.
A simple (and low risk) way would be to insert a buffer between your driver circuit and the LED driver, that has the same output specs as the circuit te LED driver was originally designed for. This would keep the on/off timing intact.
You seem to prefer to improve the performance of the driver by adding a pull-down resistor. The problem here is that for this to be effective the pull-down must have a low value compared to the series resistor R2 (which is only 1k), and that it must be sufficiency high in order to not affect the voltage level for a logic 1. I don't think there is a value that satisfies both constraints. But I guess your current R2 value is way lower than it could be, so let's say swapping the values might be a good start. (better: calculate the maximum value for R2, choose R1 at 1/10, check if that does not affect the high level).
I would prefer a third approach: change the LED driver so that it can work with an input that does not reach down to 0V. A diode in the emitter lead might be sufficient: it raises the required base voltage by ~ 0.6V.
I don't know the voltage required by your LED, if it is not too high you might get by without the CCR with placing a suitable resistor in the emitter lead, thus turning the transistor into a constant current source. An extra resistor as Olin and Dave suggest might help if the LED needs more than 2 V.