Expanding on the answer by John Snow (+1), the gist behind the bootstrap is that, if we can't go back and sample from the original distribution, the next-best option is to use the empirical distribution and sample from it. This makes the assumption that, loosely speaking, the empirical distribution is a good representation of the original distribution.
Your empirical distribution diverges from a $\text{Poisson}(2)$ distribution in a number of ways. Let $X$ be a random variable distributed as your empirical distribution.
$P(X>6) = 0$
$P(6>X>3) = 0$
For $Y\sim\text{Poisson}(2)$, it is the case that $P(Y>6) = 0.0166$ and $P(6>Y>3) = 0.1263$ (you can calculate this in R via 1 - sum(dpois(seq(0, 5), 2)) and sum(dpois(c(4, 5), 2)).
Overall, since you are sampling from a distribution that is not $\text{Poisson}(2)$, your sampling distribution does not look like it should, and since you are sampling from a distribution that has fairly marked differences from a $\text{Poisson}(2)$, your sampling distribution looks quite wrong.
The good news is that (by the Glivenko-Cantelli theorem), as you get a large sample size, your empirical distribution tends toward the true distribution, explaining why the sampling distribution looks better when you up the sample size: the sample just looks more like a $\text{Poisson}(2)$, so anything derived from it looks more like it can from a $\text{Poisson}(2)$ distribution.
For a really extreme example, suppose you sampled from $N(0,1)$ and got only positive values, by "bad luck" as you wrote (which is really just ordinary luck, as that can happen every so often). You know that the sampling distribution of the usual mean $\bar X$ in this case is $N(0, \frac{1}{\sqrt{n}})$. However, when you run the bootstrap, you will only select positive values and will only calculate positive means, meaning that the bootstrap sampling distribution gives no probability to values below zero, despite the fact that $N(0, \frac{1}{\sqrt{n}})$ has half of its density allocated to negative numbers.
