In different books (e.g. Applied Stochastic Control of Jump Diffusions) I note that the jump-diffusion process (in one dimension) can be described as a solution of \begin{equation}\label{diffusionWithJumps} dS_t = \mu(S_{t}, t) dt + \beta(S_{t}, t)dW_t + \int_{\mathbb{R}} \phi(S_{t-}, t, z) \tilde{N}(dt, dz), \end{equation}
The special case of jump-diffusion process is a Merton's model of the form \begin{equation} S_t = s e^{(r-\frac{\sigma^2}{2})t + \sigma W_t + \sum_{i=1}^{N_t}Y_i}, \end{equation} which is a solution to \begin{equation}\label{expLevyModel} dS_t = r S_{t} dt + \sigma S_{t} dW_t + S_{t-}(Y_t-1)dN_t. \end{equation}
My question is: How should I set $\mu(S_{t-}, t)$, $\beta(S_{t-}, t)$ and other components in the first of given formula to obtain the last formula. In other words how can I go from the most general case to this particular one (Merton's model).
I come a bit late to the party and I am going to admit that my answer is probably not going to be the most formal but here is the attempt.
In the Merton model, we are looking at a compound Poisson process where $lnY \sim N(\mu_y, \sigma_y)$ and $N_t \sim Poisson(\lambda)$
If you take your equation for
$$ S_t = S_0e^{\left(r-\frac{1}{2}\sigma^2\right)t + \sigma dW_t+\sum^{N_t}_{t=1}Y_t} $$
and let's replace $W_t$ by $X$ and apply Ito for jump processes we get the following:
$$ df_t = \left(\frac{\partial f}{\partial t} +\frac{\partial f}{\partial X} + \frac{1}{2}\frac{\partial'' f}{\partial X^2} \right)dt + \frac{\partial f}{\partial X}dW_t + \left[f(S_t+Y_t) - f(S_t)\right]dN_t$$
$$\frac{\partial S}{\partial t} = (r-\sigma^2/2)S_t$$ $$\frac{\partial S}{\partial X} = \sigma S_t $$ $$\frac{\partial'' S}{\partial X^2} = \sigma^2 S_t $$ $$\left[f(S_t+Y_t) - f(S_t)\right] =S_te^{Y_t} - S_t = S_t\left(Y_t-1\right)$$
And so replacing the terms above we get: $$dS_t = S_t(r-\sigma^2/2 + \sigma^2/2)dt + S_t\sigma dW_t + S_t(Y_t-1)dN_t $$ Which simplifies to $$dS_t = S_trdt + S_t\sigma dW_t + S_t(Y_t-1)dN_t $$
and so you have that $\mu(S_{t-},t) = rS_t$, $\beta(S_{t-},t) = \sigma S_t$ and $\int_\mathbb{R} (S_{t-},t,z)\tilde N(dt,dz) =S_t(Y_t-1)dN_t$ which I believe it can also be written as $\prod^{dN_t}_{j=1} \left(Y_j -1 \right)$ because $S_t$ already incorporates all the jumps up to time $t$