You are mostly correct in your reasoning, but there is a little correction to be made.
It is true that power (watts) in = power (watts) out. Watts = volts x amps.
To make the math simple, let's assume we are firing a coil at 16 watts, and the batteries we are using are at 4v. The coil is a 1Ω coil, and the board is 100% efficient (no loss of power... Not going to happen, but it makes the math simple)
So on the output side of the board, we are using 16w (assume a 1 ohm coil, it's seeing 4v, and pulling 4a). These devices change the wattage by varying the voltage into the read resistance load. So to deliver that, the input side of the board also needs to draw 16w. If you have 1 battery, you have 4v available, and you would pull 4A. If your batteries are in series, you would have 8V available, but would only be drawing 2A. (4*4=16, 2*8=16). No matter what voltage the input is seeing, has no bearing on what the output is seeing, since we didn't change the coil.
So I believe you were on the right track, and had the right idea, but your statement assumed that the coil changed.
Now if it was a mech mod, where the batteries were directly connected to the coil, then doubling the voltage would not affect the current draw, as the wattage would double as the voltage was doubled