Here is what happens without assumptions.

Functions $p_n:=\sum_i x_i^n$ represent [power-sum symmetric polynomials](https://en.wikipedia.org/wiki/Power_sum_symmetric_polynomial). By [Newton's identities](https://en.wikipedia.org/wiki/Newton%27s_identities#Expressing_elementary_symmetric_polynomials_in_terms_of_power_sums), we have
$$E(t):=\sum_{k=0}^\infty e_k \,t^k = \exp\left(\sum_{k=1}^\infty \frac{(-1)^{k+1}p_k}{k}  \,t^k \right).$$
Knowing all $p_{2n}$ is equivalent to knowing
$$E(t)E(-t) = \exp\left(-\sum_{k=1}^\infty \frac{p_{2k}}{k} \,t^{2k} \right).$$
Hence, the question amounts to computing series $E(t)$ from the known series $E(t)E(-t)$.

Let $z:=t^2$ and $E(t) = f(z) + tg(z)$ be the sum of even and odd parts. Then
$$E(t)E(-t) = f(z)^2 - zg(z)^2 = \exp\left(-\sum_{k=1}^\infty \frac{p_{2k}}{k} \,z^k \right).$$
To solve this, we can take an *arbitrary* series $g(z)$, and then compute $f(z)$ as
$$f(z) = \sqrt{\exp\left(-\sum_{k=1}^\infty \frac{p_{2k}}{k} \,z^k \right) + zg(z)^2}.$$
Then we recover values $p_n$ from
$$\sum_{k=1}^\infty \frac{(-1)^{k+1}p_k}{k}  \,t^k = \log( f(t^2) + tg(t^2)).$$

**ADDED.** If the set $\{x_i\}$ is finite, then $E(t)$ is a polynomial and so is $E(t)E(-t)$, and the problem can be solved by factoring of the latter polynomial.