Why does $\lim\limits_{n\to\infty} e^{-n}\sum_{i=1}^{n}\frac{n^i}{i!} = \frac{1}{2}$ and not 1?

The problem with your reasoning is that the two terms, $e^{-n}$ and $\sum_{i=1}^n \frac{n^i}{i!}$, can't be analyzed separately. Notice that $e^{-n}$ approaches $0$, and the second term approaches $\infty$, so the limit of the product would be $\boldsymbol{0 \cdot \infty}$, an indeterminate form. A limit of the form $0 \cdot \infty$ might equal any real number, or might even equal $\infty$.

It may be instructive to consider a different expression where some $n$s are replaced by $m$. The following limit can be evaluated as you say (I also made the sum start from $i = 0$ for simplicity): $$ \lim_{n \to \infty} e^{-\color{red}{m}} \sum_{i=0}^{\color{blue}{n}} \frac{{\color{red}{m}}^i}{i!} = 1, $$ because it is the product of limits, $e^{-m} \cdot e^m = 1$. And if we instead take the limit as $m \to \infty$, then we get $$ \lim_{m \to \infty} e^{-m} \sum_{i=0}^n \frac{m^i}{i!} = 0, $$ because the exponential beats the polynomial, and goes to $0$. In your problem, essentially, $m$ and $n$ are both going to $\infty$ at the same time, so we might imagine that the two possible results ($0$ and $1$) are "competing"; we don't know which one will win (and it turns out that the result is $\frac12$, somewhere in the middle).

How can we show that your limit is $\frac12$? This is a difficult result; please take a look at this question for several proofs (thanks to TheSilverDoe for posting).

In that question, the summation starts from $i=0$ instead of $i=1$. However, note that we can add $e^{-n}$ to your limit and it will not change (since $\lim_{n \to \infty} e^{-n} = 0$). So this gives $$ \lim_{n \to \infty} \left(\left( e^{-n} \sum_{i=1}^n \frac{n^i}{i!} \right) + e^{-n} \right) = \lim_{n \to \infty} e^{-n} \sum_{i=\color{red}{0}}^n \frac{n^i}{i!}. $$