Analytical solutions to $E[f(X_\tau) e^{-\alpha\tau}]$
Solution 1:
As discussed in the comments [big thanks to @Thomas], the answer is trivial. The stopping time $\tau$ is defined such that $X_\tau=K$. Thus, $$E_t[f(X_\tau)e^{-\alpha\tau}]=f(K)E_t[e^{-\alpha\tau}].$$