Q:

Why does Black-Scholes equation hold on continuation region of American Option? 

Explanation for Put Option:
$$ \frac{\partial V}{\partial t}+ \mathcal{L}_{BS} (V) = 0, $$ 
where
$\mathcal{L}_{BS} (V) = \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + (r-q) S \frac{\partial V}{\partial S} - r V$
holds for $S > S_f$, where $S_f$ is contact point.  
Why does this equation hold for $S > S_f$? 
Could you give me link for proof? 
Another question is: Why do we need high-contact condition? 
Update:
Do I correctly understand that for American Put Option, if $S > S_{f}$, there is no sense to exercise at time $t<T$ (because it causes immediate loss: $-V+S-K<0$). 
So  it behaves like European Option, hence $V^{Am}_{P}=V^{E}_{P}$ and it satisfies Black-Scholes Equation. 

A:

The actual problem one solves for American options is an optimal stopping time problem, so the value of the option is
$$
V_0 = \max_\tau E_{\tau}\left[e^{-r \tau} (S_\tau-K)^+  \right]
$$
where the maximum is taken over all stopping times (exercise strategies $\tau>0$ permissible in the contract). 
With a PDE operator such as you have, the instantaneous equality can be expressed in linear complementarity form as
$$ \left(\frac{\partial V}{\partial t}+ \mathcal{L}_{BS} (V)\right)\cdot \left(V-g\right) = 0  $$
where $g$ represents early exercise value. 
Note for convenience that (post exercise) the stock itself satisfies the BS PDE trivially.