Let $E$ be a normed $\mathbb R$-vector space, $(X_t)_{t\ge0}$ be a càdlàg Lévy process on a filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0},\operatorname P)$, $B\in\mathcal B(E)$ with $0\not\in\overline B$, $\tau_0:=0$ and $$\tau_n:=\inf\{t>\tau_{n-1}:\Delta X_t\in B\}$$ for $n\in\mathbb N$.
I would like to show $(\tau_n-\tau_{n-1})_{n\in\mathbb N}$ is independent and identically distributed.
We can show that $\tau_1$ is exponentially distributed with parameter $c:=-\ln\operatorname P\left[\tau_1>1\right]$.
Let $$\tilde X_t:=X_{\tau^B_1+t}-X_{\tau^B_1}$$ and $$\tilde F_t:=\mathcal F_{\tau^B_1+t}$$ for $t\ge0$.
Unless I'm missing something, $\{\tau_1^B<\infty\}$ should have probability $1$ and $(\tilde X_t)_{t\ge0}$ and $(\tilde{\mathcal F}_t)_{t\ge0}$ are at least well-defined on the trace $(\tilde\Omega,\tilde{\mathcal A})$ of $(\Omega,\mathcal A)$ on $\{\tau_1^B<\infty\}$.
On this trace, by the strong Markov property, $(\tilde X_t)_{t\ge0}$ is an $(\tilde{\mathcal F}_t)_{t\ge0}$-Lévy process with $\tilde X\sim X$.
We can clearly write $$\tau_2=\tau_1+\underbrace{\inf\{t>\tau_{n-1}:\Delta\tilde X_t\in B\}}_{=:\:\tilde\tau_1}\tag1.$$
Since $\tilde X\sim X$, $$\tilde\tau_1\sim\tau_1\tag2.$$
However, the expression $\tau_2-\tau_1$ is only well-defined on $\{\tau_2<\infty\}$. Does this set again have probability $1$? It clearly holds $\tau_n\rightarrow{n\to\infty}\infty$. So, how do we need to deal with the possibility of encountering $\{\tau_n=\infty\}$? And how do we actually need to phrase the result? Claiming that "$(\tau_n-\tau_{n-1})_{n\in\mathbb N}$ is independent and identically distributed" doesn't seem to make sense, if this process is not even well-defined (at least on $(\Omega,\mathcal A)$) ...