How do JBC calculate the power rating of their soldering stations?
If the heater resistance increases by, say 15% with temperature and the transformer voltage drops by 15%  under full load, and there is maybe 1V drop in the switching transistor and lead wires, we have:
P = \$\frac{(23.5V\cdot 0.85-1V)^2}{2.5\Omega\cdot 1.15)}\approx\$  125W
Close enough for government work.