Consider the discrete time stochastic process $m_j=\sum_{i=1}^jx_i$ for a finite time interval $j=1,\dots,n$, where the increments $x_i\sim\text{Pois}(\lambda_i)$ are all independent Poisson random variates with respective parameters $\lambda_i>0$. Therefore for every fixed $j$: $$ m_j\sim\text{Pois}(\Lambda_j),\quad\Lambda_j=\sum_{i=1}^j\lambda_i $$
Now define the first passage time to level $s$ as $$ \tau_s=\min\{j\in \{1,2,\dots,n,\infty\}\,:\,m_j\ge s\},\quad s=0,1,2,\dots $$ where, as usual, $\infty$ is included to cover the case when the process $m_j$ does not reach level $s$ until time $n$.
Is it possible to derive the distribution of $\tau_s$ in closed form?
If yes, how?
Using @qervert's great suggestion I am answering my own question:
For $s=0$ the definition implies $\tau_0=1$ on all paths of $m_j$. Hence $\mathbb{P}(\tau_0=j)=\delta_{j,1}$, where $\delta_{i,j}$ is the Kronecker symbol, and the CDF is $$ \mathbb{P}(\tau_0\le j)=1,\quad j=1,\dots,n,\infty.\tag{1} $$
For $s\ge1$ the suggestion in the comment works and the CDF for finite times is $$ \mathbb{P}(\tau_s\le j)=1-\frac{\Gamma(s,\Lambda_j)}{\Gamma(s)},\quad j=1,\dots,n,\tag{2a} $$ and we have a finite probability of not reaching level $s$ until time $n$ given by $$ \mathbb{P}(\tau_s=\infty)=\frac{\Gamma(s,\Lambda_n)}{\Gamma(s)},\tag{2b} $$ which is the missing weight needed to make a proper CDF, $\mathbb{P}(\tau_s\le\infty)=1$.