What is unit of time average of product of two signals?
The spectrum of the product and the product of the spectra are completely different things.
Let's look at a simple example. Let's look at a cirucit with a voltage source and single impedance $Z$. For simplicity we will use the complex admittance $Y = 1/Z = Y_m \cdot e^{j \varphi}$ where $Y_m$ is the magnitude and $\varphi$ is the phase of the admittance.
Let's drive this with a voltage $u(t) = u_0 \cdot cos(\omega t)$. The current waveform will then be
$$i(t) = u_0 \cdot Y_m \cdot cos(\omega t + \varphi) $$
The instantaneous power is
$$p(t) = u(t) \cdot i(t) = u_0^2 \cdot Y_m\cdot [cos(\omega t) \cdot cos(\omega t + \varphi)]$$
Using the trig identitiy we get
$$p(t) = u_0^2\cdot Y_m/2 \cdot [cos(\varphi)+cos(2 \cdot \omega \cdot t + \varphi)]$$
The time averaged power would simply be
$$&lt;p(t)&gt; = u_0^2\cdot Y_m/2 \cdot cos(\varphi)$$
The spectrum of the instantaneous power has two frequency compontents: One at $0Hz$ and one at $2 \omega$. The $0Hz$ component is the time averaged power and the actual physical power that would be dissipated in the impedance. It obviously depends on the phase of the impedance. For example for a capacitor or inductor, the dissipated power is 0 and the instantaneous power spectrum ONLY has a reactive componet at $2 \cdot \omega$.
Note that the spectrum of the instantaneous power is a VERY different from the power spectrum of both voltage and the current , i.e. only one component at $\omega$
The assumption that you can get the spectrum of the power my muliplying the spectra of the field quantities is just plain wrong. The math doesn't work out either: multiplication in the time domain is CONVOLUTION in the frequency domain, not multiplication. If you convolve the units work out as well
$$P(\omega ) = \int_{s}^{}U(s) \cdot I(s-\omega)ds$$
which has the correct units of $W/Hz$.