Why the spectral coherence is unity for all frequencies between single-frequency time series and itself
According to the documentation, the coherence between two signals $x(t)$ and $y(t)$ is defined as $C_{xy}(w) = \frac{|P_{xy}(w)|^2}{P_x(w) P_y(w)}$, where $P_x(w)$ and $P_y(w)$ are power spectral density estimates of $x$ and $y$ and $P_{xy}(w)$ is an estimate of the cross-spectral density. 


In your first experiment, you put in two identical signals $x(t) = y(t) = \cos(2\pi w_1 t)$. Since $x(t)=y(t)$, what you should expect is that $P_x (w)= P_y(w) = P_{xy}(w)$ for all $w$. Therefore, $C_{xy} = \frac{|P_{xy}(w)|^2}{P_x(w) P_y(w)}= 1$ for all $w$. 
You might want to argue that the signal(s) only contains a single frequency. In this case we would expect that $P_x(w) =P_y(w) = P_{xy}(w) =0$ for $w \neq w_1$. However:


This would mean that $C_{xy}(w) = \frac{0}{0} = $undefined for $w\neq w_1$.
This wouldn't happen numerically, as due to rounding errors you'd have some epsilons left that could still make this fraction equal to one
You're not dividing power densities but (Welch) estimates of the power densities of a finite window of the signal, hence $P(w)$ will not be exactly zero anywhere. It will contain estimation artifacts and the truncation effects. Since $x(t)$ and $y(t)$ are exactly equal, the artifacts are exactly equal and the coherence is one.

When you add noise to $x(t)$ and $y(t)$ in your second experiment you're adding "signal content" to the frequencies outside $w=w_1$. Since the noise contributions are drawn independently, they are uncorrelated which means that $P_{xy}(w)$ will indeed be zero there, while $P_x(w)$ as well as $P_y(w)$ will not be. Hence the coherence does go to zero.