this is certainly not the answer that you are looking for , but with a 720 W power input ( 12V * 60 A ) you can't get 1000 W on the output of your inverter .
Anyway you can certainly do what you want with a comparator to compare the voltage drop between two points in the main conductor that links the battery to the inverter .
If there's around 50 or 60 A flooding out of your battery, I think you won't need a shunt resistor to detect the voltage drop, there will certainly be a voltage drop across the main conductor .
When the current exceeds a certain limit, the voltage drop across this conductor could be used with a rail-to-rail comparator to generate a high level .
This output could light a led through a resistor.