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[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 $\phi(t)-\psi(t)=o(|t|)$.

[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 $\phi(t)-\psi(t)=\sum_{k=1}^N(\sum_i x_i^k - \sum_i y_i^k)t^k/k!$ 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 $\phi(t)-\psi(t)=o(|t|)$.

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

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, ans since $\phi(t) -\psi(t) =o(|t|)$ for $t\to-\infty$, it is identically zero, so $\sum_i x_i^k=\sum_i y_i^k$ for $1\le k\le N$ too. $\sum y_i <\infty$ and $\sum_i x_i^k=\sum y_i^k$ for all $k> N$ and $\psi(t):=\sum_i(e^{y_it}-1)$

[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 $\phi(t)-\psi(t)=o(|t|)$.

[incomplete] Assuming $x_i\ge0$ and $y_i\ge0$ with $ \sum_i x_i <\infty$, $\sum y_i <\infty$ and $\sum_i x_i^k=\sum y_i^k$ for all $k> N$ we have that $\phi(t):=\sum_i(e^{x_it}-1)=\sum_{k\ge1} \big(\sum_i x_i^k\big)t^k/k!$ and $\psi(t):=\sum_i(e^{y_it}-1)$ are entire functions (we can expand the exponentials and exchange order of summation, by absolute summability). It remains to prove that $\phi(t) -\psi(t) =o(|t|)$ for $t\to-\infty$; then since $\psi(t)-\phi(t)=\sum_{k=1}^N(\sum_i x_i^k - \sum_i y_i^k)t^k/k!$ is a polynomial, it would follow it is identically zero, so $\sum_i x_i^k=\sum_i y_i^k$ for $1\le k\le N$ too.

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

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, ans since $\phi(t) -\psi(t) =o(|t|)$ for $t\to-\infty$, it is identically zero, so $\sum_i x_i^k=\sum_i y_i^k$ for $1\le k\le N$ too. $\sum y_i <\infty$ and $\sum_i x_i^k=\sum y_i^k$ for all $k> N$ and $\psi(t):=\sum_i(e^{y_it}-1)$