@LocalVolatility proves in this stellar answer that European call option prices in the Merton jump diffusion model are given by $$ C_{Merton}(S_0,r,q,\sigma,K,T) = \sum_{n=0}^\infty e^{-\lambda T}\frac{(\lambda T)^n}{n!} C_{BS}(S_0^{(n)},r,q,\sigma^{(n)},K,T). $$
Question: Does the same reasoning apply to American options, where we take the maximum over all stopping times into account? Is the price of American call (put) options in the Merton model equal to the sum of Black-Scholes American calls (puts), weighted by the probability of $n$ jumps occurring? Do we need to adjust $S_0$ and $\sigma$ for all $n$ in the same way as for European calls (puts)?
The argument, based on the law of iterated expectation, could begin like this \begin{align*} c_{American}&=\max\limits_{\tau\;stopping\;time\;on\;[0,T]}\mathbb{E}^\mathbb{Q}[e^{-r\tau}(S_\tau-K)^+] \\ &=\max\limits_{\tau\;stopping\;time\;on\;[0,T]}\mathbb{E}^\mathbb{Q}\big[e^{-r\tau}\mathbb{E}^\mathbb{Q}[(S_\tau-K)^+|N_\tau=n]\big] \\ &= \max\limits_{\tau\;stopping\;time\;on\;[0,T]} \sum_{n=0}^\infty \mathbb{Q}[N_\tau=n] \mathbb{E}^\mathbb{Q}[e^{-r\tau}(S_\tau-K)^+|N_\tau=n] \\ &= \max\limits_{\tau\;stopping\;time\;on\;[0,T]} \sum_{n=0}^\infty e^{-\lambda \tau}\frac{(\lambda \tau)^n}{n!} \mathbb{E}^\mathbb{Q}[e^{-r\tau}(S_\tau-K)^+|N_\tau=n]\\ \end{align*} The latter part looks like a standard American call option in the BS world (with the same adjustments to $S_0$ and $\sigma$ as for European options) but how to deal with the jump probabilities that depend on the stopping time $\tau$?
