There is an interesting thought in this question: knowing the power consumption is enough to determine voltage and current.
There is indeed only one solution at a given power consumption for discrete components. A 1\$\Omega\$ resistor will consume 1W when it is running at 1V and 1A. For a diode, the operating point can be found on the IV curves even if it is not straightforward.
For a capacitor we can also determine what voltage it is at for a given energy storage which would be equal to \$P.t\$ where t is time in seconds, and the instantaneous current can be calculated as \$P/V\$.
Using power for analysing circuit behavior can be a powerfull technique (no pun intended). It can help you find the amplitude of an output signal faster than more complex computations.
However, it is not very handy to use power as the unique or even the main specification of these components. For a diode for instance, the voltage level hardly changes for a big current range because the VI curve is almost vertical after the 1.x V knick. And also because the component can operator at many power levels, and currents, but mostly at a "single" voltage level. So for a diode, the operating voltage is a major characteristic.
Once we know the voltage we still have to limit the power consumption. If I say that the LED should operate at 30mW is that a convenient information to work with? No! When I have a known drive voltage, I want to determine the resistor value to determine the set point for the LED. The LED power does not allow me to determine it directly. Voltage and current do! I just have to know how much voltage the resistor needs to handle and how much current I want. If my source is 5V, my LED operates at 2V, I have 3V over the resistor, so knowing that I want about 20mA, I know the resistor's value by doing 3x50= 150\$\Omega\$ (because I know that 50 = 1000/20).
If I were using power, I'ld have 40mW, so I'ld need to convert that to voltage and current, or at least know the voltage. With voltage and power alone, I would use the fact that the resistor consumes 1.5 times the power of the LED because it has 1.5 times the voltage drop. So it consumes 60mW. \$P=V.I=V^2/R\$ so \$R=V^2/P\$ meaning that \$R=9V^2/60mW=9\Omega/0.06=3\Omega/0.02=150\Omega\$. Wow, the same result, but a bit harder to come by.
So rather than computing the interesting characteristics each time, we memorize them or communicate them. It speeds up our work, and keeps us concentrated on the important stuff. Power consumption becomes interesting to know too because in the our world of connected devices, determinining the power budget is important too, but is is less important if you just want the circuit to work.