If we let $S$ be the set of characters of $\mathbb F_p^\times$ trivial on $G$ then $$\sum_{\chi \in S} \chi(g) = \begin{cases} \frac{p-1}{|G|} & g\in G \\ 0 & g\notin G \end{cases}$$

so $$\sum_{\chi_1,\chi_2\in S} \sum_{ g \in \mathbb F_p \setminus \{0,1,2\} } \chi_1(g) \chi_2(2-g)  = 0$$ if $G$ is almost trivial.

Now if $\chi_1 =\chi_2=1$ then $\sum_{ g \in \mathbb F_p \setminus \{0,1,2\} } \chi_1(g) \chi_2(2-g)=p-3$ and otherwise, by the bound for Jacobi sums, $\left| \sum_{ g \in \mathbb F_p \setminus \{0,1,2\} } \chi_1(g) \chi_2(2-g) \right| \leq  \sqrt{p}+1 $.

Thus, if $G$ is almost trivial then $p-1 \geq ( ((p-1)/|G|)^2-1) (\sqrt{p}-1)$,  or $p \geq c^2 p^{5/2} / |G|^2$ for a constant $c$, meaning $|G| \leq c p^{3/4}$, answering the weaker form of your question.