Here is an observation which does not solve your problem, but translates it into one which has a large literature. Note that the $k$-th elementary symmetric function can be expressed as a polynomial with rational coefficients in the first $k$ power symmetric functions. For example, $e_2(x_i) = \Sigma(x_ix_j | i < j) = 1/2[(\Sigma(x_i))^2 - \Sigma(x_i^2)] = 1/2[(p_1(x_i))^2 - p_2(x_i)]$. This implies in your problem that if $Q(z) = \Pi (z - x_i)^{r_i} = z^a + q_1z^{a-1} + q_2z^{a-2} + \cdots $ (where $a = r_1 + \cdots + r_k$), and $S(z) = \Pi(z - y_i) = z^n + s_1z^{n-1} + s_2z^{n-2} + \cdots $, then $q_i = s_i$ for $ i = 1, \dots , m+n-1$. Equivalently, $Q(z) - z^{a-n}S(z)$ has degree $\leq a - (m+n)$. This is a situation where the much generalized Mason-Stothers Theorem applies.

For a polynomial $F(z)$ over $\mathbb{C}$, let $h(F)$ be the number of distinct roots of $F$, and let $d(F)$ be the degree of $F$. Mason-Stothers Theorem: If $Q(z) + T(Z) + V(z) = 0$, where $Q,T,V$ are complex polynomials with no common root, then $h(Q) + h(T) + h(V) \geq max(d(Q),d(T),d(V)) + 1$.

The theorem was first proved in W. W. Stothers, Polynomial Identities and Hauptmoduln, Quart. J. Math. Oxford 32 (1981), 349-370. It can be proved in a few lines by elementary means, but most of Stothers' paper uses deeper methods from the theory of automorphic forms to study the boundary case where $h(Q) + h(T) + h(V) = max(d(Q),d(T),d(V)) + 1$. In the boundary case he gives examples of choices of root multiplicities for $Q,T,V$ for which there are solutions and examples for which there are no solutions. I couldn't tell if his examples include any of your polynomial systems. Since his method requires looking at finite index subgroups of a free group of rank 2, I expect that more examples could be analyzed with today's computers.

When there is a solution to one of your polynomial systems, let $Q(z),S(z)$ be as in the first paragraph above, and let $T(z) = -z^{a-n}S(z), V(z) = -Q(z) + z^{a-n}S(z) = -Q(z) - T(z)$. Then $h(Q) = m$, $h(T) = n + 1$, $a - (m +n) \geq d(V) \geq h(V)$, $d(Q) = a$, and Mason-Stothers implies that $ a+1 = m + (n + 1) + (a - (m + n)) \geq h(Q) + h(T) + h(V) \geq d(Q) + 1 = a + 1$, so all inequalities become equalities, which is the boundary case.

Here is an observation which does not solve your problem, but translates it into one which has a large literature. Note that the $k$-th elementary symmetric function can be expressed as a polynomial with rational coefficients in the first $k$ power symmetric functions. For example, $e_2(x_i) = \Sigma(x_ix_j | i < j) = 1/2[(\Sigma(x_i))^2 - \Sigma(x_i^2)] = 1/2[(p_1(x_i))^2 - p_2(x_i)]$. This implies in your problem that if $Q(z) = \Pi (z - x_i)^{r_i} = z^a + q_1z^{a-1} + q_2z^{a-2} + \cdots $ (where $a = r_1 + \cdots + r_k$), and $S(z) = \Pi(z - y_i) = z^n + s_1z^{n-1} + s_2z^{n-2} + \cdots $, then $q_i = s_i$ for $ i = 1, \dots , m+n-1$. Equivalently, $Q(z) - z^{a-n}S(z)$ has degree $\leq a - (m+n)$. This is a situation where the much generalized Mason-Stothers Theorem applies.

For a polynomial $F(z)$ over $\mathbb{C}$, let $h(F)$ be the number of distinct roots of $F$, and let $d(F)$ be the degree of $F$. Mason-Stothers Theorem: If $Q(z) + T(z) + V(z) = 0$, where $Q,T,V$ are complex polynomials with no common root, then $h(Q) + h(T) + h(V) \geq max(d(Q),d(T),d(V)) + 1$.

The theorem was first proved in W. W. Stothers, Polynomial Identities and Hauptmoduln, Quart. J. Math. Oxford 32 (1981), 349-370. It can be proved in a few lines by elementary means, but most of Stothers' paper uses deeper methods from the theory of automorphic forms to study the boundary case where $h(Q) + h(T) + h(V) = max(d(Q),d(T),d(V)) + 1$. In the boundary case he gives examples of choices of root multiplicities for $Q,T,V$ for which there are solutions and examples for which there are no solutions. I couldn't tell if his examples include any of your polynomial systems. Since his method requires looking at finite index subgroups of a free group of rank 2, I expect that more examples could be analyzed with today's computers.

When there is a solution to one of your polynomial systems, let $Q(z),S(z)$ be as in the first paragraph above, and let $T(z) = -z^{a-n}S(z), V(z) = -Q(z) + z^{a-n}S(z) = -Q(z) - T(z)$. Then $h(Q) = m$, $h(T) = n + 1$, $a - (m +n) \geq d(V) \geq h(V)$, $d(Q) = a$, and Mason-Stothers implies that $ a+1 = m + (n + 1) + (a - (m + n)) \geq h(Q) + h(T) + h(V) \geq d(Q) + 1 = a + 1$, so all inequalities become equalities, which is the boundary case.