No, it isn’t. But all formulas only work if the variables are fixed (and this might be fixed for a specific situation). For Ohms law (R = U / I), you can only calculate R is U and I are fixed (for that situation).
And your calculations are true, but it isn’t the same coffee maker*
Let’s make an analogy. You have a car and you know the average km/l (MPG). Now you’re going to make a journey and you know the length, you can calculate the fuel consumption. But that is, if you drive the same as you always do and the road is average aka if you may assume your km/l is valid. But if that road is 10% up and you drive twice as fast you will consume more fuel. But you can g up that hill with the same km/l as normal, but your average speed probably is a lot lower
Same goes for loads. Something is fixed. For a heating element (and roughly for an incandescent) R is fixed. Which means that only leaves one variable in Ohm’s law. Or you can define the voltage and calculate the current (I = U / R) or you can define the current and calculate the voltage drop you get (U = I x R). But more complex loads don’t have R fixed. The model can be A LOT more complex than just a resistor. But let’s assume we have a load with 100W @ 12V and 100W @ 230V. Which means the power now is fixed. And accrding to Ohm’s law and the power law, that means the other parts are not fixed. You can define the voltage which leads to a specified R (R = U2 / P) and I (I = P / U). Or you can define a current and calculate U (U = P / I) and R (R = P / I2). As you can see, in both cases all the right hand variables are or fixed or you defined them.
But it all depends on the load you have. Seeing every load as a resistor just isn’t true And a resistor isn’t the only load with resistance, it’s simply a part/model with fixed resistance (unless otherwise specified like a PTC or LDR).
*At least not a normal one because they really are not fixed power but fixed resistance (roughly) If a coffee maker is designed for 1000W @ 230V it will not be