Let $f\in C_0$, the space of all continuous functions vanishing at infinity. Let $A\subset \mathbb{R}$ be such that $\text{span}(f_x \mid x\in \bar{A})$ is dense in $(C_0,\|\cdot\|_{\infty})$, where $f_x:\mathbb{R}\to \mathbb{R}$ defined by $f_x(y)=f(xy)$.
Is it true that $\text{span}(f_x \mid x\in A) $ also dense in $(C_0,\|\cdot\|_{\infty})$?
It suffices to show that for any $x \in \bar{A}$ there exists $(h_n)_{n \in \mathbb{N}} \subset \text{span}(f_x \mid x \in A)$ such that $\|h_n-f_x\|_{\infty} \to 0$ as $n \to \infty$.
To this end, we first observe that the denseness of $\text{span}(f_x \mid x \in \bar{A})$ in $C_0$ implies that $f(0) \neq 0$. Indeed, if $f(0)$ were zero, then $h(0)=0$ for any $h \in \text{span}(f_x \mid x \in \bar{A})$ which would mean that the family cannot be dense in $C_0$ (just pick some $g \in C_0$ with $g(0) \neq 0$).
Now let $x \in \bar{A}$ and $\epsilon>0$. Since $f_0(y)=f(0) \notin C_0$, we have $x \neq 0$. By definition, there exists $(x_n)_{n \in \mathbb{N}} \subset A$ such that $x_n \to x$. Moreover, $f \in C_0$ implies that we cann choose $R>0$ such that
$$|f(uy)| \leq \epsilon \quad \text{for all $|u| \geq \frac{|x|}{2}$, $|y| \geq R$}.\tag{1}$$
As $f$ is uniformly continuous we can also choose $\delta>0$ such that
$$|f(uy)-f(vy)| \leq \epsilon \quad \text{for all $|u-v| \leq \delta$, $|y| \leq R$}. \tag{2}$$
Since $x_n \to x$ and $x \neq 0$, it holds that $|x_n| \geq |x|/2$ and $|x_n-x| \leq \delta$ for $n \geq N$ sufficiently large. On the one hand, we have by $(1)$
$$|f_x(y)-f_{x_n}(y)| \leq |f(xy)|+ |f(x_ny)| \leq 2 \epsilon$$
for all $|y| \geq R$ and $n \geq N$; on the other hand, by $(2)$
$$|f_x(y)-f_{x_n}(y)| =|f(xy)-f(x_ny)| \leq \epsilon$$
for all $|y| \leq R$ and $n \geq N$. Hence,
$$\|f_x-f_{x_n}\|_{\infty} \leq 2 \epsilon, \qquad n \geq N.$$
As $\epsilon>0$ is arbitrary, this proves that we can find $(h_n)_{n \in \mathbb{N}} \subset \text{span}(f_x \mid x \in A)$ such that $\|h_n-f_x\|_{\infty} \to 0$ as $n \to \infty$.