As I assume you don't want to buy new drivers, you have one option: add another relay.

If the drivers can run without any load attached to them, just disconnect the leds before you cut ac power and add a bleed resistor before the 2nd relay (10k 1/2 watt or thereabouts). The datasheet should be very clear on this.

If the driver cannot tolerate this, use the relay in combination with a power resistor to 'short' the output right before you turn of the ac power. Also this should only happen at very low brightness levels to avoid fire. The resistance should be chosen as such that the driver's output voltage always stays below the minimum led voltage at maximum possible current.

Example:

of leds: 10

led minimum voltage: 2.8V
voltage margin: 0.3V
minimum turn on voltage for the chain: 28V
maximum current the driver can supply: 1A

--> R = ( 28V - 3V ) / 1A = 25 Ohm (most likely you'll only find 22 Ohm).

wattage: 25W

But as the resistor only has to burn energy for a very short time, I guess you could get away with a 7W resistor as well. Nevertheless it should be in a well ventilated area an not be near any easily inflammable material. It would also be a good idea to add a suitable (~1A or less) slow-blow fluse in series with the resistor, just in case something goes wrong with the logic.