The joint distribution off all order statistics $X_{(1)}, \dots, X_{(n)}$ is
$$f_o(y_1, \dots, y_n) = n! f(y_1) \times \cdots \times f(y_n)$$
for $y_1 \le \cdots \le y_n$.
Thus, the joint distribution of $X_1 ,\dots, X_n$, given $X_{(1)}, \dots, X_{(n)}$ does not depend of the density $f$! We have
$$ Pr(X_1 = x_1, \dots, X_n= x_n | X_{(1)} = y_1, \dots, X_{(n)} = y_n ) = {1\over n!} $$
when the multisets of the $x_i$'s and of the $y_i$'s are equal.
The intuition is that loosing the drawing order doesn’t matter, for the $X_i$'s are independent.
PS Re-reading this answer long after, I tend to think that the conclusion is clear in itself: given $y_1 < \dots < y_n$, $Pr(X_1 = x_1, \dots, X_n= x_n | X_{(1)} = y_1, \dots, X_{(n)} = y_n ) = {1\over n!}$ means that all re-orderings of the $y_i$'s are equally probable, whatever the density $f$ is. In fact I don't see how to prove the starting result I used (the joint distribution of order statistics) without proving this at the same time... This just comes from the fact that all the points $x = (x_1, \dots, x_n)$ obtained from a permutation of the $y_i$ have the same density $\prod_i f(x_i) = \prod_i f(y_i)$.
Best Answer
A sufficient statistic is sufficient for a particular family of probability distributions, and in this case that family is actually not the family of Poisson distributions, each member of which is supported on the set $\{0,1,2,3,\ldots\},$ but rather it is a family of distributions supported on the set $\{0,1,2,3,\ldots\}^n$ of all $n$-tuples of finite cardinal numbers. Thus the random variable involved is an $n$-tuple $(X_1,\ldots,X_n)$ in which the components are independent and each has a $\operatorname{Poisson}(\lambda)$ distribution.
Within the set of all such $n$-tuples one finds the observable data $(x_1,\ldots,x_n)$. If the observable data are altered within that set, the number $n$ does not change; only the $x\text{s}$ change. Thus the proof is not erroneous.
If the number of independent identically distribued Poisson observations were a random variable in its own right, then one would have a different family of distributions and the sum of the observations would probably not be a sufficient statistic.