if $X_1,\ldots,X_n$ are independent with $X_i \sim \mathrm{Pois}(1/n)$ (and so $E(X_i) = \operatorname{Var}(X_i)=1/n)$ then $S_n=\sum\limits_{i=1}^n X_i\sim \mathrm{Pois}(1)$ for all $n$.
If the CLT holds, it follows that $S_n\sim \mathrm{Pois}(1)$ is a normal
distribution
No, you're misapplying the theorem.
Suppose $n=6$. Then you have $X_1,\ldots,X_6 \sim \mathrm{i.i.d. Poisson}\left(\dfrac 1 6 \right)$.
Then $X_1+\cdots+X_6 \sim \mathrm{Poisson}(1)$. And $X_1+\cdots+X_{600} \sim \mathrm{Poisson}(100) \ne \mathrm{Poisson}(1)$. And $\mathrm{Poisson}(100)$ is approximately normal with expectation $100$ and variance $100$ (so the standard deviation is $10$). But use a continuity correction here: if you want $\Pr(X_1+\cdots+X_{600}>95)$, realize that that is the same as $\Pr(X_1+\cdots+X_{600}\ge96)$ and so plug in $95.5$.
If you then change the distribution of $X_1,\ldots,X_{600}$ to $\mathrm{Poisson}\left(\dfrac 1 {600}\right)$, then their sum is indeed distributed as $\mathrm{Poisson}(1)$, which is not so close to normal. But if $X_1,\ldots,X_{6000000}\sim\mathrm{i.i.d. Poisson}\left(\dfrac 1 {600}\right)$ then again you get something that is approximately normal.
The central limit theorem says if $X_1,X_2,X_3,\ldots\sim\mathrm{i.i.d.}$ with a finite variance, then the distribution of
$$
\frac{(X_1+\cdots+X_n)-\operatorname{E}(X_1+\cdots+X_n)}{\operatorname{SD}(X_1+\cdots+X_n)}
$$
approaches the standard normal distribution as $n\to\infty$. It does not say that if you keep changing the distribution of each of the random variables as $n$ increases, then the same thing happens.