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Let $n$ be the number of users who want to access a channel divided into with $s$ time-slots and let $n<<s$.
Each user randomly chooses one time-slot (out of $s$) for accessing the channel.
A collision occurs if more than one user chooses a given time-slot.

The probability of successful access for everybody without any collisions ( considering $n<<s$) can be approximated by $e^{-n^2/2s}$.
The expected number of users who experience a collision is also given by $n(1-(1-1/s)^{n-1})$.

This part of the problem is actually equivalent to the birthday problem where a year is assumed to have $s$ days.

Now, assume that in the case of a collision in the first round $i=1$ (time-slots $1,..., s$), the users who experience a collision will back off and try to access the channel, again randomly, during the next round $i = 2$ with $s$ time-slots (time-slots in $[s+1, 2s]$) and this procedure goes on during the next $s$ time-slots, if a collision occurs, until everybody accesses the channel.

The question is, how many rounds are required on average so that all the users access the channel once?
What would be a closed-form approximation?

The expected number of people who experience collisions in a round is
$$n(1 - (1 - 1/s)^{n-1}) approx n(1 - (1 - n/s)) = n^2 / s$$
as you noted. So we expect the number of people $P_i$ still participating after $i$ rounds to satisfy the recurrence
$$P_0 = n qquad P_{i+1} = frac{P_i^2}{s}$$
The first few values are $n, frac{n^2}{s}, frac{n^4}{s^3}, frac{n^8}{s^7}, cdots$ and the closed form is
$$P_i = frac{n^{2^i}}{s^{2^i-1}}$$
Solving for when this is less than 1,
$$n^{2^i} = s^{2^i - 1}$$
$$2^i log n = (2^i - 1)log s$$
$$log s = 2^i(log s - log n)$$
$$log log s = i + log(log s - log n)$$
$$i = log log s - log(log s - log n) = log(1 + frac{log n}{log s - log n})$$
which is approximately $log log n$ assuming $n << s$.

This is just a heuristic estimation of the expected number of rounds in total.