Given $P \div z^{n-k} $ evaluated in the roots of $P$, find $P$. Say there is some polynomial $P=(z-a_1)(z-a_2)\cdots(z-a_n) = Q(z) z^{n-k} + R(z)$ where the degree of $R$ is less than $n-k$.
Now say we are given $q_1=Q(a_1), q_2=Q(a_2), \dots, q_n=Q(a_n)$, but not the roots themselves.
Can we recover $P$ and the roots $a_1, a_2, \dots, a_n$?
If we expand $P$ with the elementary  symmetric polynomials and insert the roots, we get a polynomial system with $n$ equations and $n$ unknowns, so I assume it's solvable somehow.
Of course if $k=0$ we just have $Q(z) =1$, but say $k=1$, then $Q(z) =z-\sum_i a_i$.
Let $s=\sum_i a_i$, then
$a_i-s=q_i$.
Summing we get $s-ns=\sum_i q_i$.
So we can recover $s$ and then the roots.
 A: The answer is already NO for $(n,k)=(4,2)$. Indeed, for that case the formulas are
$$
q_1 = a_2a_3+a_2a_4+a_3a_4, 
q_2 = a_1a_3+a_1a_4+a_3a_4, \ \textrm{etc}.
$$
Notice that each $q_j$ is homogeneous with degree $2$. So all those $q_j$ stay unchanged if $(a_1,a_2,a_3,a_4)$ is replaced by $(-a_1,-a_2,-a_3,-a_4)$. In particular, (unless all the $a_k$'s are all zero) there are at least two distinct polynomials $P$ that will yield the same $(q_1,q_2,q_3,q_4)$.
In general, the $q_j$'s will be homonegenous of degree $k$, and therefore invariant wrt multiplication by a $k$-th root of unity.
UPDATE 08/18/2022  A little more can be said. It is straightforward to check that $q_j$ equals $(-1)^k$ times the $k$-th symmetric polynomial in all the $a$-variables except $a_j$.
In particular, if $\lambda$ and $t$ are two parameters, $l$ is an index in $[1..n]$ and we take $a_j=t$ for $j\neq l$ and $a_j=-\lambda t$ for $j=l$, then
$$
q_j=\begin{cases} \bigg(\binom{n-2}{k}-\binom{n-2}{k-1}\lambda\bigg)t^k, \ \textrm{if} \ j \neq l, \\ 
\binom{n-1}{k} t^k, \ \textrm{if} \ j = l.
\end{cases}\tag{1}
$$
If we choose $\lambda = \frac{\binom{n-2}{k}}{\binom{n-2}{k-1}}$, we then have
$$
q_j=\begin{cases} 0 \ \textrm{if} \ j \neq l, \\ 
\binom{n-1}{k} t^k, \ \textrm{if} \ j = l.
\end{cases}\tag{2}
$$
This shows that $q_l$ is algebraically independent from $\lbrace q_j \rbrace_{j\neq l}$. So the extension ${\mathbb Q}(q_1,q_2,\ldots,q_n)$ of $\mathbb Q$ has transcendance degree exactly $n$, as does the extension ${\mathbb Q}(s_1,s_2,\ldots,s_n)$, where $s_j$ denotes the $a$-th symmetric polynomial in all the $a$-variables.
In particular, each $s_j$ is algebraic over ${\mathbb Q}(q_1,q_2,\ldots,q_n)$, so that there are only finitely many $P$'s yielding a given $(q_1,q_2,\ldots,q_n)$.
When $(n,k)=(4,2)$, with the help of a computer I found that the minimal polynomial of $s_1=a_1+a_2+a_3+a_4$ over ${\mathbb Q}(q_1,q_2,\ldots,q_n)$ is an atrociously complicated polynomial with degree $8$ in $s_1$ and degree $20$ in each of $q_1,q_2,\ldots,q_4$.
This suggests that the general problem might be very hard, unless there is a clever trick I missed.
