Let $q=2^h$ and $t=2^r$ for some $h\ge r$ and denote by $\mathbb{F}_q$ the finite field of order $q$.
(since the previous, simple version was wrong, I'm posting here a new version)
Let $f$ be a monic polynomial with $f(0)=0$, $f(1)=0$, such that the equation $f(z)=c$ has either $0$ or $t$ solutions in $\mathbb{F}_q$. Let $F_a(z)=\frac{f(z)+f(a)}{z+a}$, and we are given that the restriction of $F_a(z)$ to its non-roots (and $\neq a$) is injective, and there are constants $w_1,...,w_{t-1}$ such that none of the polynomials $w_i^{-1}+F_a(z)$ has a root in $\mathbb{F}_q\setminus\{a\}$.
Computationally, for $q\le 32$, I turns out that $\{0,w_1,\ldots,w_{t-1}\}$ is always an additive subgroup in $\mathbb{F}_q$, and that the $\{z|f(z)=c\}$ are also cosets of an additive subgroup. I wonder if any of this could be proven in general.
If $t\neq 1,2,q$, then there exist subsets of $\mathbb{F}_q$ of size $t$ that include $0$, but that are not additive subgroups. For example, take $S = (\mathbb{F}_t\setminus \{1\}) \cup \{a\}$, where $a\notin \mathbb{F}_t$.
Since any function from $\mathbb{F}_q$ to itself can be interpolated with a polynomial, this is quite false as stated—just let $f$ be any $t$-to-$1$ function with $f(S) = 0$—though I'm still curious what the idea is, and whether there is a less trivial statement of the problem.