The sum of the Poisson distribution is equal to exactly 1 only when the sum of $P(X = j)$ is taken for all values of j (0 to infinity):
$$\sum_{j=0}^{\infty} e^{-\lambda}\frac{\lambda^j}{j!} = 1$$
When using a Poisson distribution to approximate a Bernoulli trial where an experiment with probability of success $p$ is performed a finite number $n$ times, I think it only makes sense for $j$ to go up to $n$, so the sum of the probabilities of all possible outcomes should be
$$\sum_{j=0}^{n} e^{-\lambda}\frac{\lambda^j}{j!}$$
which is less than 1. Does this mean that if we use the Poisson distribution to approximate a Bernoulli trial then we should expect that the sum of the probability of all possible outcomes be less than one by $\sum_{j=n+1}^{\infty} e^{-\lambda}\frac{\lambda^j}{j!}?$
You should expect the sum off all the exact probabilities to equal one. As you argue, the sum of the Poisson-approximated probabilities for zero through $n$ will be less than one. If we are in a parameter range where the Poisson approximation is accurate (i.e. large $n,$ small $p$, with $np$ normal sized), then this should not matter very much and the sum will be very close to one. The impossible results greater than $n$ that the approximating Poisson erroneously thinks are possible will have a very small probability.