Given a birth process $\{B_t:t\geqslant0\}$ with $\lambda >0$, define $$K_t=\int_{0}^{t}B_s ds=\sum_{i=1}^{n}B_{t_{i}}(t_{i+1}-t_i)$$ if there were $n$ births in $[0,t]$ and let $t_{i}$ be the times of birth. Note that the soujorn time $$S_{i}:=(t_{i+1}-t_i)\sim\mathrm{ Exp}(\lambda) .$$ Define $$ \sigma_{t}=\inf\{s\geq 0:K_{s}>t\} $$ Can we think of the time scaling $t\rightarrow \sigma_t$ that $B_{\sigma_t}$ is Poisson?
The reason behind this:
It is well known, that $$ B_{t}-\underbrace{\lambda\int_{0}^{t}B_{s}ds}_{A_t} $$ is a Martingale, where $A_{t}$ is a strictly increasing continuous process . Hence $A_t$ is the compensator of $B_t$, which is a counting process, since $A_0=0$ the representation is unique. Since $A_t$ is continuous, $B_t$ is quasi-left-continuous. For a quasi-left-continuous counting process $B_t$ with compensator $A_t$ define the stopping time $$ \tau_t:=\inf\{s\geq 0:A_s>t\} $$ Then $B_{\tau_t}$ is poisson process with rate 1, see Theorem 5 here. The relation is $\lambda K_t=A_t$. As Did mentions, it holds $$ \sigma_{t}=\tau_{\lambda t}. $$
There are some problems considering the starting value $B_{0}=x$, since a poisson-prozess starts at 0. So can we think of $B_{\sigma_t}$ as a Poisson-Prozess starting at $x$?
If this is not valid, then we just consider $\tilde{B_t}=B_{t}-B_{0}$ (which is the classical counting process starting at 0, which counts the births) and the martingale is $$ B_{t}-B_{0}-\lambda \int_{0}^{t}B_sds $$ With the same argumentation as above and same defined stopping time $\tau_{t}$ then least $\tilde{B}_{\tau_{t}}$ is a poisson process with rate 1. Since $\tilde{B}_{\lambda \tau_t}=\tilde{B}_{\sigma_{t}}$, our process $(\tilde{B}_{\sigma_{t}})_{t\geq 0}$ is a poisson-process with rate $\lambda$.