Fix a sequence $(a_n)_{n=0}^\infty$ of positive real numbers satisfying $\sqrt[n]{a_n}\to 0$, so that the rule $F(x)=\sum_{i=0}^\infty a_ix^i$ defines a function $F:\mathbb{R}\to\mathbb{R}$.
Conjecture. There exist $N\in\mathbb{N}$, a sequence $(b_n)_{n=0}^\infty$ of nonnegative real numbers, and a sequence $(x_i)_{i=1}^\infty$ of real numbers satisfying the following.
(i) $0\leq b_n\leq a_n$ for all $n\geq N$;
(ii) $b_n>0$ for some (equivalently, infinitely many) $n$;
(iii) $|x_i|\to\infty$; and
(iv) $G(x_i)=0$ for all $i$, where $G:\mathbb{R}\to\mathbb{R}$ is defined by the rule $G(x)=\sum_{n=0}^\infty b_nx^n$.
Discussion.
This is motivated by an observed consequence of Picard's Great Theorem. Indeed, if we define $F_\mathbb{C}:\mathbb{C}\to\mathbb{C}$ by $F_\mathbb{C}(z)=\sum_{n=0}^\infty a_nz^n$ (so that $F_\mathbb{C}\equiv F$ on $\mathbb{R}$) then the map $z\mapsto F_\mathbb{C}(1/z)$ has an essential singularity at zero. By Picard's Great Theorem, there is a negative number $d\in(-\infty,0)$ such that $F_\mathbb{C}(1/z)=d$ for infinitely many $z\in\mathbb{C}$. Equivalently, $F_\mathbb{C}(z_i)=d$ for a sequence $(z_i)_{i=1}^\infty$ of distinct complex numbers. If we define $b_0=a_0-d$ and $b_n=a_n$ for $n\geq 1$, we get $G_\mathbb{C}(z_i)=0$ all $i$, where $G_\mathbb{C}:\mathbb{C}\to\mathbb{C}$ is defined by the rule $G_\mathbb{C}(z)=\sum_{n=0}^\infty b_nz^n$. As $G_\mathbb{C}$ is entire, we must have $|z_i|\to\infty$.
However, I need the $x_i$'s for which $G(x_i)=0$ to be real. As the $b_n$'s are all nonnegative, this means in particular that we are looking at $x_i$'s in $(-\infty,0)$. The condition that $b_n>0$ for infinitely many $n$ ensures that $G$ is nonconstant.
Proof idea.
Here is an example to show that this is sometimes possible. If $F(x)=e^x$ then we can define $G(x)=\frac{1}{2}[e^x+\sin(x)]$ to get what we need.
Mimicking the above example, one strategy would be to:
(1) Find an auxiliary function $H(x)=\sum_{n=0}^\infty c_nx^n$ with $0<c_n\leq a_n$ and $\lim_{x\to-\infty}H(x)=0$; and
(2) Try to show that the map $J(x)=\sum_{n=0}^\infty(-1)^nc_{2n+1}x^{2n+1}$ satisfies $\limsup_{x\to-\infty}J(x)>0$ and $\liminf_{x\to-\infty}<0$.
Then set $b_{2n+1}=\frac{1}{2}[1+(-1)^n]c_{2n+1}$ and $b_{2n}=\frac{1}{2}c_{2n}$ for all $n$. Clearly, this satisfies (i) and (ii) in the conjecture. Furthermore, the conditions in (1) and (2) guarantee that the function $G(x)=\sum_{n=0}^\infty b_nx^n=\frac{1}{2}[H(x)+J(x)]$ oscillates, thus satisfying (iii) and (iv).
However, I am not currently able to get (1) or (2). And although (1) seems plausible, (2) does not.
For any sequence of positive reals $a_n$, we can define $b_n > 0$ inductively. Let $P_n(x) = \sum_{j=0}^{2n+1} b_j x^j$. Suppose $r_n$ is the most negative real root of $P_n$. Thus for some $\epsilon_n > 0$ we have $P_n(r_n) = 0$, $P_n(r_n - \epsilon_n) < 0$, $P_n(r_n + \epsilon_n) > 0$. With appropriate bounds on $b_j$ for $j > 2n+1$, we can ensure that this will still be true for $P_m$ where $m > n$. Now given $b_0, \ldots, b_{2n+1}$, take $b_{2n+2} > 0$ (small enough to satisfy the bounds obtained previously). Now $P_n(x) + b_{2n+2} x^{2n+2} \to \infty$ as $x \to -\infty$: take some $c$ so this is positive at $x=c$. Then add $b_{2n+3}$ where $b_{2n+3} > 0$ is small enough so $P_{n+1}(x) = P_n(x) + b_{2n+2} x^{2n+2} + b_{2n+3} x^{2n+3}$ still has $P_{n+1}(c) > 0$, but $P_{n+1}(x) \to -\infty$ as $x \to -\infty$. Then take $r_{n+1}$ so $P_{n+1}(r) = 0$.
$G(z) = \sum_{j} b_j z^n = \lim_{n \to \infty} P_n(z)$ is analytic (if the $b_j$ go to $0$ fast enough, and by construction we have $G(r_n - \epsilon_n) \le 0 \le G(r_n + \epsilon_n)$, so $G$ has a root in $[r_n - \epsilon_n, r_n + \epsilon_n]$.