You're right. I think the article's reasoning is fallacious. At least, I don't see what the Poission distribution has to do with this at all.

One way to arrive at the answer correctly is to write the number of samples needed before they first sum to 1 as N and observe that the probability that P[N > k] = 1/k!. You can get this from a k-dimensional integral. A general formula for E[N] is Sum_k=0^infty P[N > k] thus E[N] = e.