I do not understand the question, because it starts with a wrong statement:

> A superposition of two electromagnetic waves with different frequencies will never produce visible interference patterns. 

This statement is clearly false. Take, for example, the diffraction pattern from a double slit with a single frequency:

[![Double slit diffraction pattern with a single wavelength][1]][1]

If you take a wavelength that is 10% longer you get a slightly wider spacing:

[![Double slit diffraction pattern with a longer wavelength][2]][2]

When you superpose the two wavelength, the intensity that you get is simply the sum of the two. The result still has a very clear diffraction pattern, showing interference:

[![Double slit diffraction pattern with two wavelengths][3]][3]

It is easy to prove that the intensities of the two waves with different wavelength are simply added. Let us say that the first wave is:
$$ \varphi_1 = \psi_1(x) e^{i \omega_1 t} $$
and the second:
$$ \varphi_2 = \psi_2(x) e^{i \omega_2 t} $$
The square modulus of the superposition is:
$$ |\varphi|^2 = \left| \varphi_1 + \varphi_2 \right|^2 $$
Substituting:
$$ |\varphi|^2 = \left| 
\psi_1(x) e^{i \omega_1 t} + 
\psi_2(x) e^{i \omega_2 t} \right|^2
$$
This is equal to:
$$ |\varphi|^2 = \left(
\psi_1(x) e^{i \omega_1 t} + 
\psi_2(x) e^{i \omega_2 t} \right)
\left(
\psi_1(x) e^{i \omega_1 t} + 
\psi_2(x) e^{i \omega_2 t} \right)^*
$$
The four terms are:
$$ |\varphi|^2 = 
\psi_1(x) e^{i \omega_1 t} \psi_1(x)^* e^{-i \omega_1 t}
+
\psi_1(x) e^{i \omega_1 t} \psi_2(x)^* e^{-i \omega_2 t}
+ 
\psi_2(x) e^{i \omega_2 t} \psi_1(x)^* e^{-i \omega_1 t}
+
\psi_2(x) e^{i \omega_2 t} \psi_2(x)^* e^{-i \omega_2 t}
$$
Some of the exponential cancel out:
$$ |\varphi|^2 =
\psi_1(x) \psi_1(x)^* + \psi_2(x) \psi_2(x)^*
+ 2 Re \left( 
\psi_1(x) \psi_2(x)^* e^{i (\omega_1-\omega_2) t}  
\right)
$$
Calculating the intensity, the last term on the right averages to 0:
$$ I =
\psi_1(x) \psi_1(x)^* + \psi_2(x) \psi_2(x)^*
= I_1 + I_2
$$

So, from the superposition of two diffraction patterns at different wavelengths you simply get the sum of the intensities. Normally, the diffraction will still be visible.






  [1]: https://i.sstatic.net/oApzM.jpg
  [2]: https://i.sstatic.net/ddhdb.jpg
  [3]: https://i.sstatic.net/W6I6I.jpg