For your first question, here's a simple proof.

- Observe that $\partial_{ii}^2 e_k = 0$ for any $i\in \{1,\ldots,n\}$. This implies
$$ \partial^2_{ii} (\sigma^k f) = 0 \tag{1}$$
for any $i$, where $\sigma = \sum x_i$.
- Observe that
$$ \partial_i \sigma = 1 $$
for any $i$, and hence
$$ \partial_i \sigma^k = k \sigma^{k-1} \tag{2}$$
by the chain rule.
- The product rule applied to (1) using (2) gives
$$ 0 = k(k-1)\sigma^{k-2} f + 2 k \sigma^{k-1} \partial_i f + \sigma^k \partial^2_{ii} f$$
from which we divide by $\sigma^k$ to obtain
$$ \partial^2_{ii} f + k(k-1)\sigma^{-2} f + 2 k \sigma^{-1} \partial_i f = 0 \tag{3}$$
Recall that this holds
*individually for any $i$*.
- Multiply now the expression (3) by $x_i$. And sum over $i$, we get finally
$$ \sum_{i = 1}^n x_i \partial^2_{ii} f + k(k-1) \sigma^{-2} f \underbrace{\sum x_i}_{= \sigma} + 2 k \sigma^{-1} \sum_{i = 1}^{n} x_i \partial_i f = 0 \tag{4} $$
- Now observe that $f$ is, by definition, a homogeneous function of degree 0. So the homogeneous derivative $\sum x_i \partial_i f = 0$. This kills the last term in (4). The remainder is exactly what you wanted to show.

Concerning uniqueness, observe that only two ingredients were used in the derivation of the PDE above:

- That the original function $e_k$ satisfies $\partial^2_{ii} e_k = 0$ for any $i$.
- That the original function $e_k$ is homogeneous of degree $k$.

There are a lot more functions that satisfy the same property. For example, when $n = 3$ and $k = 2$, we know that it works for $e_k = x_1 x_2 + x_2 x_3 + x_3 x_1$. But the same also works for $x_1 x_2$ or $x_2 x_3$ or $x_3 x_1$ or any linear combination thereof.