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

1. 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 $\sigma = \sum x_i$.
2. 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.
3. 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 $i$.
4. Multiply now the expression (3) by $x_i$. And sum over $i$, we get finally $$ \sum_{i = 1}^n x_i \partial^2_{ii} f + k(k-1) \sigma^{-2} f \underbrace{\sum x_i}_{= \sigma} + 2 k \sigma^{-1} \sum_{i = 1}^{n} x_i \partial_i f = 0 \tag{4} $$
5. 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.

For your second question: the PDE is type changing: depending on orthant, the PDE can be elliptic, hyperbolic, or ultrahyperbolic. And the coefficients degenerate along hyperplanes of the form $\{x_i = 0\}$, where the equation is formally parabolic. So on Euclidean space it is a bit hard to see how to apply the usual theory of PDE. But a naive guess is that without additional constraints there is little reason to believe that the solution is unique.

One possible way to approach this, however, is to restrict to, for example, solutions which are degree zero homogeneous. This would allow you to project down to one lower dimension. And you can further restrict to an orthant where the type of the equation is fixed.

As an example, one avenue of approach is to consider the set $\Sigma$ containing all points for which $\sum x_i = 1$ and $x_i \in (0,1)$. Under the degree zero homogeneous assumption the equation (4) should reduce to a elliptic PDE on $\Sigma$ (with coefficients degenerating near the boundary, possibly). Then it is conceivable that $f$ is generated by the unique solution to that PDE with Dirichlet boundary conditions. But this is more computations than I want to do here now.

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

1. 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 $\sigma = \sum x_i$.
2. 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.
3. 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 $i$.
4. Multiply now the expression (3) by $x_i$. And sum over $i$, we get finally $$ \sum_{i = 1}^n x_i \partial^2_{ii} f + k(k-1) \sigma^{-2} f \underbrace{\sum x_i}_{= \sigma} + 2 k \sigma^{-1} \sum_{i = 1}^{n} x_i \partial_i f = 0 \tag{4} $$
5. 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:

1. That the original function $e_k$ satisfies $\partial^2_{ii} e_k = 0$ for any $i$.
2. 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.

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

1. 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 $\sigma = \sum x_i$.
2. 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.
3. 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 $i$.
4. Multiply now the expression (3) by $x_i$. And sum over $i$, we get finally $$ \sum_{i = 1}^n x_i \partial^2_{ii} f + k(k-1) \sigma^{-2} f \underbrace{\sum x_i}_{= \sigma} + 2 k \sigma^{-1} \sum_{i = 1}^{n} x_i \partial_i f = 0 \tag{4} $$
5. 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 \tag{4). The remainder is exactly what you wanted to show.

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

1. 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 $\sigma = \sum x_i$.
2. 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.
3. 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 $i$.
4. Multiply now the expression (3) by $x_i$. And sum over $i$, we get finally $$ \sum_{i = 1}^n x_i \partial^2_{ii} f + k(k-1) \sigma^{-2} f \underbrace{\sum x_i}_{= \sigma} + 2 k \sigma^{-1} \sum_{i = 1}^{n} x_i \partial_i f = 0 \tag{4} $$
5. 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.

For your second question: the PDE is type changing: depending on orthant, the PDE can be elliptic, hyperbolic, or ultrahyperbolic. And the coefficients degenerate along hyperplanes of the form $\{x_i = 0\}$, where the equation is formally parabolic. So on Euclidean space it is a bit hard to see how to apply the usual theory of PDE. But a naive guess is that without additional constraints there is little reason to believe that the solution is unique.

One possible way to approach this, however, is to restrict to, for example, solutions which are degree zero homogeneous. This would allow you to project down to one lower dimension. And you can further restrict to an orthant where the type of the equation is fixed.

As an example, one avenue of approach is to consider the set $\Sigma$ containing all points for which $\sum x_i = 1$ and $x_i \in (0,1)$. Under the degree zero homogeneous assumption the equation (4) should reduce to a elliptic PDE on $\Sigma$ (with coefficients degenerating near the boundary, possibly). Then it is conceivable that $f$ is generated by the unique solution to that PDE with Dirichlet boundary conditions. But this is more computations than I want to do here now.