Let $\mu$ be the reference measure of the Poisson process. By definition, this means that the number of arrivals falling into any $n$ mutually disjoint sets $B_1,\ldots,B_n$ has the same distribution as independent Poisson random variables with means $\mu(B_1),\ldots,\mu(B_n)$.
Specializing to the case of a Poisson process on $[0,\infty)$ (so that the interarrival times are well-defined) we see that if $T$ denotes the first arrival, then
$$\mathbb P(T>t)=\mathbb P(0\text{ arrivals in }[0,t])=e^{-\mu([0,t])},$$
where the last equality uses the probability of a Poisson variable with mean $\mu([0,t])$ being zero.
You can see from this that if $\mu$ is not a constant multiple of Lebesgue measure, then $T$ is no longer an exponential random variable.
The interarrival times are no longer independent in general, since if $T'$ denotes the next interarrival time after $T$ then
$$
\mathbb P(T'>t\mid T)=e^{-\mu([T,T+t])}.
$$
Whenever $\mu$ is not a constant multiple of Lebesgue measure, this conditional probability depends on $T$, and thus $(T,T')$ are not independent.
The joint distribution of $(T,T')$ satisfies
$$
\mathbb P(T'>t,T>s)=\mathbb E[e^{-\mu([T,T+t])};T>s],
$$
you can not go too much farther without knowing more about the reference measure $\mu$.
Kallenberg's textbook is a good reference for Poisson process questions.
It is interesting that process has independent increments, but interarrival times are dependent. – tern Aug 12 '19 at 09:16