You didn't answer all my questions so I'm not sure whether this will help.
Connecting a load in the cathode path of an SCR is not normal. Normally, the cathode is connected to the 0V (negative) rail of the circuit, and the load is connected between a positive voltage and the anode. I am going to assume that you can rearrange your circuit into this topology.
I've done a bit of Googling and discovered that normal SCRs cannot be turned off by bringing the gate negative relative to the cathode. A special type, called Gate Turn-Off (GTO) SCRs, can be turned OFF in that way, but your 100A device is not one of these. One of the documents I found said (I think) that GTO SCRs are limited to about 10A current ratings. Since your load only draws 2A continuous, and its cold current isn't likely to exceed 10A, you may be able to use a smaller SCR that can be turned OFF via its gate (i.e. a GTO SCR).
Assuming you stick with a standard SCR, in a DC circuit where there is no automatic commutation due to the supply dropping to zero on each half-cycle, the normal way to turn an SCR off is to force its anode negative relative to its cathode for long enough for it to stop conducting. I think the obvious way would be using a capacitor connected to the anode. The other side of the capacitor would be connected through a resistor to the positive supply, or perhaps a somewhat lower positive voltage. When the SCR is ON, the capacitor would charge up to that voltage (minus the SCR's anode-to-cathode voltage drop).
To turn the SCR OFF, the side of the capacitor that's connected to the resistor to the positive voltage needs to be brought quickly and firmly to 0V. By an N-channel MOSFET, presumably. When that happens, the SCR's anode will be pulled negative. The capacitor needs to be large enough that the SCR's anode will stay negative for long enough for the SCR to turn OFF, while the anode is being pulled positive by the load. If you can guarantee a certain minimum ON-time for the SCR, you may be able to assume that the bulb has warmed up and the load current has dropped to around 2A when you come to turn the SCR OFF.
The resistor from the capacitor to the positive voltage needs to be low enough to charge the capacitor fully (or nearly fully) within the shortest time that the SCR will be ON for.
I hope this helps.