The question is, I argue, more interesting if thought about more generally, letting the distribution of the parent Poisson depend on n, say with parameter λn and λn=1 as a special case. I think it's perfectly reasonable to ask why, and how we can understand that, a central limit theorem does not hold for the sum Sn=∑ni=1Xi,n. After all, it's common to apply a CLT even in problems where the distributions of the components of the sum depend on n. It's also common to decompose Poisson distributions as the distribution of a sum of Poisson variables, and then apply a CLT.
The key issue as I see it is that your construction implies the distribution of Xi,n depends on n in such a way that the parameter of the distribution of Sn does not grow in n. If you would instead have taken, for example, Sn∼Poi(n) and made the same decomposition, the standard CLT would apply. In fact, one can think of many decompositions of a Poi(λn) distribution that allows for application of a CLT.
The Lindeberg-Feller Central Limit Theorem for triangular arrays is often used to examine convergence of such sums. As you point out, Sn∼Poi(1) for all n, so Sn cannot be asymptotically normal. Still, examining the Lindeberg-Feller condition sheds some light on when decomposing a Poisson into a sum may lead to progress.
A version of the theorem may be found in these notes by Hunter. Let s2n=Var(Sn). The Lindeberg-Feller condition is that, ∀ϵ>0:
1s2n∑i=1nE[Xi,n−1/n]2I(|Xi,n−1/n|>ϵsn)→0,n→∞
Now, for the case at hand, the variance of the terms in the sum is dying off so quickly in n that sn=1 for every n. For fixed n, we also have that the Xi,n are iid. Thus, the condition is equivalent to
nE[X1,n−1/n]2I(|X1,n−1/n|>ϵ)→0.
But, for small ϵ and large n,
nE[X1,n−1/n]2I(|X1,n−1/n|>ϵ)>nϵ2P(X1,n>0)=ϵ2n[1−e−1/n]=ϵ2n[1−(1−1/n+o(1/n))]=ϵ2+o(1),
which does not approach zero. Thus, the condition fails to hold. Again, this is as expected since we already know the exact distribution of Sn for every n, but going through these calculations gives some indications of why it fails: if the variance didn't die off as quickly in n you could have the condition hold.