Let $A$ be commutative ring and $G$ a finite group of ring automorphisms of $A$. We have a finite ring extension $A^G \to A$ that induces a surjective map $\phi: Spec(A) \to Spec(A^G)$. I'd like to prove that given $a \in A$ and $b_i \in A^G$ such that $$t^n+\sum_{i \leq n} b_i t^i=\prod_{\sigma \in G}(t-\sigma(a)) ,$$ we have $\phi(D(a))= \cup D(b_i)$ where $D(f)$ is the canonical principal open subset.
I managed to prove one inclusion (namely $\subset$) but I've got no idea for the other one.
Take $A=\mathbb{Z}$, $G=1$.
Then $5^3 = 125 = 3\times 25 + 10 \times 5 = 3\times 5^2 + 10\times 5$; $3,10\in A^G$; but $D(5)\neq D(3)\cup D(10)$ (note that $\phi = id$). So you need some more conditions for it to work, if it can work.
With the new condition : For $i=0$, $b_0 = \displaystyle\prod_{\sigma\in G}\sigma(a)$, thus if $p$ is a prime ideal of $A^G$ and $\phi(q) = p$, and $a\in q$, then $b_0 \in q\cap A^G = p$, thus $p\notin D(b_0)$. Thus $p\in D(b_0)\implies q\in D(a) \implies p\in D(\phi(a))$.
More generally, if $G= \{id=\sigma_0, \sigma_1,...,\sigma_n\}$ we have $b_i = \displaystyle\sum_{i_0<...<i_l} \prod_{k=0}^l \sigma_{i_k}(a)$ for $l=n-i$. Thus if $p$ is prime in $A^G, p=q\cap A^G$ with $q$ prime in $A$, and $b_i\notin p$, then $b_i\notin q$, so for some $i_0<...<i_l$, $\displaystyle\prod_{k=0}^l \sigma_{i_k}(a)\notin q$. Applying $g=\sigma_{i_0}^{-1}$ to both sides we get in particular $a\notin g\cdot q$. But $\phi(g\cdot q) = \phi(q) = p$, so $g\cdot q \in D(a)$ and $p=\phi(g\cdot q)\in \phi(D(a))$, thus $D(b_i)\subset \phi(D(a))$.
Thus $\cup_i D(b_i)\subset \phi(D(a))$. The other inclusion is easy as you pointed out.