I think your approach is generally correct. I will note that for your first phase, so long as $A(t) < \epsilon n$, the process $A(t)$ is well approximated by a branching process where the offspring distribution is 1 + Poisson (1). Since this has mean equal to 2 and cannot get extinct, the Kesten-Stigum theorem says that at time $t$, $A(t)$ really is of order $W 2^t$ for some random variable $W>0$. You see indeed that it thus takes $t =\log_2 n $ to make this equal to $n$, so phase 1 takes $\log_2 n $ + a random variable whose tail probabilities are uniformly bounded in $n$, as needed.

In phase 2, you can either use the ODE approach with Ethier-Kurtz type of arguments, as suggested by QAMS, or simply note the following. The probability that a Binomial $(N,p)$ deviates from its mean $Np$ (say, is less than $Np/2$) is exponential in $Np$. This means that, over a logarithmic number of trials, the probability you would observe one such deviations tends to 0. Hence during phase 2, you know that each step you add at least a $(N-A(t))\epsilon/2$ individuals, which shows that phase 2 indeed only takes a constant number of steps with overwhelming probability.

Phase 3 is a bit more delicate (you want to avoid a coupon-collector effect where collecting the last individual takes more time than it should), but I think this sort of reasoning should help you get started...