I think you are pretty much on the right track. Here is my favorite way to look at the link between Poisson and exponential distributions:
Suppose the number $N_t$ of random events that occur in time interval $(0, t]$, where $t > 0,$ has the distribution $Pois(\lambda t).$ Thus, the probability
of seeing no events by time $t > 0$ is $P(N_t = 0) = e^{-\lambda t}.$
[Notice that when dealing with a Poisson distribution over a time interval,
it is necessary to coordinate the rate $\lambda t$ with the length of the
interval.]
Another way to specify that there are no events in the interval $(0, t]$ is to let $W$ be the waiting time, starting at time $t = 0,$ until we see the
first event. Then
$$P(W > t) = 1 - F_W(t) = P(N_t = 0) = e^{-\lambda t},$$
for $t > 0.$
According to the usual notation, $F_W$ is the CDF of $W.$
By differentiation of the CDF we get the density function of $W:$
$$f_W(t) = F_W^\prime(t) = \lambda e^{-\lambda t},$$
for $t > 0$ (and 0 elsewhere). We say that $W$ has an exponential
distribution with rate $\lambda.$
Using integration by parts, one can show that
$E(W) = \int_0^\infty tf_W(t)\,dt = 1/\lambda.$ [Some texts and software
parameterize the exponential distribution in terms of the rate $\lambda$
and others in terms of the mean $\mu = 1/\lambda.$]
Thus starting with the distribution of the discrete Poisson distribution
of $N_t,$ we have found the distribution of the continuous random
variable $W.$
As a specific example: let $\lambda = 2$ per minute. Then the probability of no
events within $(0, 0.5]$ minutes is $P(N = 0),$ where $N \sim Pois(1).$ In
R statistical software, this is:
dpois(0, 1) # 'dpois' is Poisson PDF
## 0.3678794
It is also $P(W > 0.5),$ where $W \sim Exp(rate\,\, \lambda = 2).$ In R we have:
1 - pexp(.5, rate=2) # 'pexp' is exponential CDF
## 0.3678794
Simulating a million realizations of $W$ and averaging them, we have a good
estimate of the exponential mean $E(W):$
w = rexp(10^6, rate=2) # 'rexp' simulates exponential realizations
mean(w); sd(w); mean(w > 0)
## 0.5003201 # aprx E(W) = 1/2
## 0.5000453 # aprx SD(W) = E(W)
## 0.368334 # aprx P(W > 0), fraction of w's exceeding 0
It is not difficult to show that an exponential random variable has
variance $1/\lambda^2 = \mu^2,$ so that $SD(W) = 1/\lambda = \mu.$
Acknowledgment: This discussion is much that same as one finds at
the beginning of Ch 4 in Suess and Trumbo (2010), Springer.
