System Identification with Periodic Signal Input
The point is, that you need to look at the influence of averaging on the final term in the equation

$$ \hat{G}_N = [...] + \frac{V_N}{U_N}. $$

In case of periodic excitation $u$, the noise term $|V_N|$ is decreased by averaging over independent realizations (=measurements). However $|U_N|$ is not affected by averaging, as each period of u(t) is identical.

In case of non-periodic excitation, both terms are affected in the same way. As each is different, $|U_N|$ is reduced by averaging and so does the quotient $|\frac{V_N}{U_N}|$ and  $|E_N|$.

Note: In your example plots you do not take averages in the non-periodic case. The frequency resolution in the first plot is different for each N. If the non-periodic signal is split in segments with equal length and averaged over segments, the frequency resolution would be the same for all N. However in your plot  you just compute longer DFTs that result in finer frequency resolution. However this is not relevant for the answer.

Edit: Extended my notes on DFT lenghts; thanks to Maximilian Matthé for his comment.