If $x(t)$ is a finite-energy signal with Fourier transform $X(f)$, then $x(t)\cos(2\pi f_c t)$ is also a finite-energy signal with Fourier transform $\left.\left.\frac 12 \right[X(f-f_c) + X(f+f_c)\right]$. This is just the modulation theorem of Fourier transform theory.
The energy spectral density of $x(t)$ is $S_x(f) = |X(f)|^2$ while the energy spectral density of $y(t) = x(t)\cos(2\pi f_c t)$ is
\begin{align}S_y(f) &= \bigg|\left.\left.\frac 12 \right[X(f-f_c) + X(f+f_c)\right]\bigg|^2\\
&= \left.\left.\frac 14 \right[ |X(f-f_c)|^2 + |X(f+f_c)|^2 + 2\Re\big(X(f-f_c)X^*(f+f_c)\big)\right]\\
&= \left.\left.\frac 14 \right[S_x(f-f_c)+S_x(f+f_c)+2\Re\big(X(f-f_c)X^*(f+f_c)\big)\right]\tag{1}
\end{align}
which has a vague resemblance to what the OP wrote. Now, if $x(t)$ is a low-pass signal with bandwidth $W$ Hz and $W < f_c$, then $X(f-f_c)$ occupies the frequency band from $f_c-W > 0$ to $f_c+W$ while $X(f+f_c)$ occupies the frequency band from $-f_c-W$ to $-f_c+W < 0$. We see that there is no value of $f$ for which $X(f-f_c)$ and $X(f+f_c)$ both are nonzero, and thus $X(f-f_c)X^*(f+f_c)$ equals $0$ for all $f$. We are left with
\begin{align}
S_y(f) &= \left.\left.\frac 14 \right[S_x(f-f_c)+S_x(f+f_c)\right]\\
&= \left.\left.\frac 14\right[|X(f-f_c)|^2+|X(f+f_c)|^2\right].\tag{2}\end{align}
But when $f_c$ is smaller than $W$, then, as the OP peevishly has informed us all that he has placed no restrictions on the value of $f_c$, he must learn to live with the more complicated $(1)$.
