if LED always "eats" 2 V on every current, other 7 V will have to be dissipated on internal resistance. current will rise until IxR_internal=7 V. on battery terminals you will actually measure 2 V, same as on diode terminals.What happens when I pull the resistor leaving only the L.E.D. across the 9V battery? I want to calculate both amperage & resistance, but it's a little confusing. I know there's a 2V drop across the L.E.D., but I also know at those same points there'll be a 9V drop across the terminals of the battery.
this will be a large current which will destroy your LED in short time, LED will not be able to cool itself.
your battery schematics is "9 V source + internal resistance". battery resistance is serial resistor which can not be removed.