= Solution
First take $k$ to be a nonnegative integer. The <exponential tilting> constructed from the independent increments restricts to $\mathbb P_n^\lambda$ on each $\mathcal F_n$. Since its <Radon-Nikodym derivative> is strictly positive, finite-time change of measure gives
$$
\mathbb P(\tau_k\leq n)=\mathbb E^\lambda[(M_n^\lambda)^{-1}\mathbf1_{\{\tau_k\leq n\}}].
$$
Under the tilted law, $L_n=(M_n^\lambda)^{-1}$ is a <martingale>, because
$$
\mathbb E^\lambda[\phi(\lambda)e^{-\lambda X_{n+1}}]=\int\phi(\lambda)e^{-\lambda x}\frac{e^{\lambda x}}{\phi(\lambda)}\,\mu(dx)=1.
$$
Use the <stopped likelihood ratio under exponential tilting> at this finite horizon. In detail, $\{\tau_k=j\}\in\mathcal F_j$, so
$$
\begin{aligned}
\mathbb E^\lambda[L_n\mathbf1_{\{\tau_k\leq n\}}]
&=\sum_{j=0}^n\mathbb E^\lambda[\mathbb E^\lambda(L_n\mid\mathcal F_j)\mathbf1_{\{\tau_k=j\}}]\\
&=\sum_{j=0}^n\mathbb E^\lambda[L_j\mathbf1_{\{\tau_k=j\}}]
=\mathbb E^\lambda[L_{\tau_k}\mathbf1_{\{\tau_k\leq n\}}].
\end{aligned}
$$
Here the last expression is defined only on the displayed event, so no value of $M_\infty^\lambda$ is being assumed. The <upward skip-free random walk> has integer increments bounded above by one, which imply $S_{\tau_k}=k$ on a finite hit: the preceding value is at most $k-1$, and there is no upward overshoot. Consequently
$$
\boxed{\mathbb P(\tau_k\leq n)=e^{-\lambda k}\mathbb E^\lambda[\phi(\lambda)^{\tau_k}\mathbf1_{\{\tau_k\leq n\}}].}
$$
At $\lambda=\lambda_0$, the <mean under one-sided exponential tilting> is finite and positive. The <strong law of large numbers> gives $S_n/n\to\phi'(\lambda_0)>0$ under $\mathbb P^{\lambda_0}$, so every integer level is hit <almost surely>. Since $\phi(\lambda_0)=1$, passing to $n\to\infty$ gives
$$
\boxed{\mathbb P(\tau_k<\infty)=e^{-\lambda_0k},\qquad k\in\mathbb Z_{\geq0}.}
$$
Let $H=\sup_{n\geq0}S_n$. The event $\{H\geq k\}$ is exactly the finite-hit event: an integer sequence with all its values below $k$ has supremum at most $k-1$. Also $\mathbb P(H=\infty)=\lim_k e^{-\lambda_0k}=0$. Subtraction of consecutive tail probabilities gives the <geometric maximum of an upward skip-free random walk>:
$$
\boxed{\mathbb P(H=j)=(1-e^{-\lambda_0})e^{-\lambda_0j},\qquad j=0,1,2,\ldots.}
$$
This is the <geometric distribution> counting failures before the first success, with success parameter $1-e^{-\lambda_0}$.
The PDF writes only $k\geq0$, without explicitly declaring integer levels. If it is read as allowing real $k$, then $\tau_k=\tau_{\lceil k\rceil}$ and the displayed factor must be $e^{-\lambda\lceil k\rceil}$; the hitting probability is likewise $e^{-\lambda_0\lceil k\rceil}$. For example, increments $+1$ and $-1$ with probabilities $1/3$ and $2/3$ have $\lambda_0=\log2$: hitting level $1/2$ has probability $1/2$, not $2^{-1/2}$. The integer interpretation supplies the intended formula.
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