From various experiments, I seem to have rediscovered the following idea
Proposition (Uniform randomization of visits yields Poisson process). Suppose that a process iteratively visits elements in a set where each visit selects an element at equal probability (uniform distribution across the set). Let $z$ be the time it would take to scan all of the elements once (scan the set). Then from the perspective of any given element, the visits arrive as a Poisson process with rate $\zeta = 1/z$.
I say rediscovered because I suspect that the above proposition is a consequence of well-known principles of Poisson processes. However, I haven't found anything explicit. Where might I find something?
My sketch proof of the proposition: Choose a target element from the set. Let $T$ be the time to between visits to the target. It is sufficient to show that $T$ has an exponential distribution with rate parameter $\zeta$ as this occurs if and only if the target is receiving visits as a Poisson process with rate $\zeta$. Partition the set into $n$ subsets that require equal durations to scan. By construction, the target can reside in only one subset. Moreover each subset takes duration $\tfrac{d}{n}$ to scan at so $m$ visits will take duration $m\tfrac{d}{n}$. Now for any $t>0$ $$ \mathbb{P}(T \leq t) = 1 - \mathbb{P}(T > t) $$ Put $m = \tfrac{tn}{d}$ so that $m$ visits take duration $t$. Consequently $T > t$ occurs if and only if there are $m$ visits that each miss the target hence $$\begin{align*} \mathbb{P}(T \leq t) &= 1 - \mathbb{P}(\text{$m$ visits that each miss the target}) \\ &= 1 - \left( \frac{n-1}{n} \right)^m \end{align*}$$ as each visit selects a subset at equal probability and there are $n-1$ subsets that do not contain the target. Therefore $$\begin{align*} \mathbb{P}(T \leq t) &= 1 - \left( \left(1- \frac{1}{n} \right)^n \right)^{\frac{m}{n}} \\ &= 1 - \left( \left(1- \frac{1}{n} \right)^n \right)^{\zeta t} \\ &\rightarrow 1 - \exp(-\zeta t) & \text{as $n \to \infty$} \end{align*}$$ which is the cumulative distribution function for the exponential distribution with rate parameter $\zeta$, as required.
What I have rediscovered is the convergence in distribution of $X/n$ to an exponential distribution where $X$ has a geometric distribution with $p=1/n$. Each of the visits is a Bernoulli trial, and the condition of "counting the number of visits until the revisited element is selected" is equivalent to $X$.
See "The Geometric Distribution" in "Section 3.8: Convergence in Distribution" of Siegrist, Kyle (2020, August 10) Probability, Mathematical Statistics, and Stochastic Processes. Retrieved June 24, 2021, from https://stats.libretexts.org/@go/page/10148