R, 25 bytes

c(sample(99,rexp(1),T),0)

Try it online!

p(0) at each iteration is (e-1)/e.
p(each other number) at each iteration is (1/e)*(1/99).

Obviously this choice of random distribution gives a rather unsatisfying-looking output (since most of the runs are rather short). So this link uses the same approach, but changes p(0) to roughly 0.01 to illustrate some longer runs...