[*edit: completed*]  Assuming  $x_i\ge0$ with $ \sum_i x_i <\infty$,  we have that  $\phi(t):=\sum_i(e^{x_it}-1)=\sum_{k\ge1} \big(\sum_i x_i^k\big)t^k/k!$ is an entire function  (we can expand the exponentials and exchange order of summation, by absolute summability). 

Now if for an other non-negative sequence  $y_i$ we have $\sum  y_i <\infty$  and $\sum_i x_i^k=\sum y_i^k$ for all $k> N$ we consider the corresponding entire function $\psi(t):=\sum_i(e^{y_it}-1)$:
then  $\psi(t)-\phi(t)=\sum_{k=1}^N(\sum_i x_i^k - \sum_i y_i^k)t^k/k!$ is a polynomial. It is sufficient to prove that $\phi(t) -\psi(t)  =o(|t|)$ for $t\to-\infty$, and it follows it is identically zero, so $\sum_i x_i^k=\sum_i y_i^k$ for $1\le k\le N$ too.

To show $\phi(t)-\psi(t)=o(|t|)$ for (real) $t\to -\infty$:  each term $e^{x_it}-e^{y_it}$ tends to $0$ as $t\to-\infty$, and  $|e^{x_it}-e^{y_it}|\le |t||x_i-y_i|\le |t|(x_i+y_i)$ because the function $\exp$ is $1$-Lipschitz on $\mathbb R_-$. Therefore by dominated convergence of series, $(\phi(t)-\psi(t))/t\to0$ as $t\to-\infty$, that is $\phi(t)-\psi(t)=o(|t|)$.