"the current supplied remain constant and the batteries just drain less"
The LED current will be unaffected by the addition of the second identical parallel battery.
V = I x R
In this circuit you are doubling the battery, but not changing the output voltage (two identical 9V batteries in parallel is still a 9V output).
On the load side, the resistor and LED have not changed (that's the R in Ohm's law). Please note an LED is not accurately modeled as a pure resistance, but a complete explanation of that is not necessary to understand the answer to your question.
No change in V; No change in R; ...therefore NO CHANGE in I (current)
E = V x I x t
What does change is the total potential energy in this circuit. If you double the battery count, the total current sourced to the LED will be unchanged, but the current supplied by each battery will be 1/2 of the total. Because the batteries are supplying half the current as before, they will last twice as long.
Energy is voltage times current times the time the current is supplied at that voltage. A 1000mAh Alkaline battery means that it can supply 1A at ~1.4V for ~1 hour.
No change in E; No change in V; ...therefore battery life (time) is INVERSELY proportional to current