The 7809 will have no problems if you switch off the LED, no. As long as you keep the input voltage above 11V (regulator output voltage + regulator dropout voltage), it will regulate okay.
Also, you will need a heatsink if you want to draw up to 750mA.
The regulators thermal resistance θj-a is 65°C/W, so for every watt dissipated it rises 65°C above ambient temperature.
Your input voltage is 12V, and the output 9V, so at 1A, the dissipation equals:
(Vin - Vout) * Iout -> (12V - 9V) * 750mA = 2.25W
so the temperature rise above ambient without a heatsink would be:
2.25W * 65 = 146.25°C - too much, the ICs thermal protection will kick in and shut it down.
So you need a heatsink, to bring the θj-a down to at least, say 35°C/W, preferably much lower. Here's a reference on Heatsink Basics. This is an example of a heatsink that would help keep your IC cool. It's thermal resistance is 24°C/W, so added to the 5°C/W thermal resistance of the junction to case (which we can't do anything about) we get 24°C + 5°C = 29°C/W. So the temp rise above ambient is now 2.25W * 29 = 65.25.
Not great, but you have some margin there.
For your maximum ambient working temp, subtract this value from 125°C (maximum working temp for the regulator) and you get 125 - 65.25 = 59.75°C. This can easily be reached in an enclosed circuit, so usually you would want a bit more "headroom" than this, e.g. the ability to work up to 70°C ambient. There are loads of heatsinks to choose from out there though, many with very low thermal resistances that will handle this dissipation easily.