digital IQ modulators
The power of the final real-valued signal $y_2$ is half the power of $y_1$ because you take the real part (and thus reduce power by a factor 2). The real part of $y_1$ is already the sum of the I-component and the Q-component, not only the I-component:

$$Re\{y_1\}=Re\{x\}\cos\omega_1 t - Im\{x\}\sin\omega_1 t= I-Q$$

So the original formula $I^2+Q^2$ is correct, you just need to apply a factor of $1/2$ if you don't compensate for taking the real-part. This is why in some textbooks the real part operation in a bandpass modulator is 'implemented' as $\sqrt{2}Re\{\cdot\}$; this will keep the power unchanged.