Let $X$ be a compact metric space, and let $B$ be a convex balanced bounded set in $C(X)$ such that for every $x\in X$ there is $f\in B$ with $f(x)\ne 0$.
Let $M=\{u\in C(X),~ uf\in B,~\forall f\in B\}$ and let $N=\{u\in C(X),~ uf\in \overline{B},~\forall f\in \overline{B}\}$.
Since multiplication $(f,g)\to fg$ is a continuous operation on $C(X)$, it follows that $N$ is closed in $C(X)$ and $M\subset N$. Thus, $\overline{M}\subset N$.
Is it true that $N=\overline{M}$?
Let $X=[0,1]$, let $p_n(x)=x^n$ for $n\in{\mathbb N}$, let $a(x) =e^{-x}$.
Let $V_0 = \operatorname{lin}\{p_n \colon n\in {\mathbb N}\}$ and let $V = {\mathbb C}a + V_0$.
Claim 1: $V_0 = \{ f \in V \colon f(0)=0\}$.
Proof: the LHS is obviously contained in the RHS; for the converse inclusion, note that $a(0)=1$ while $p_n(0)=0$ for all $n\in{\mathbb N}$. $\qquad\Box$
Claim 2. Let $u\in C(X)$. If $up_1\in V$ and $ua\in V$ then $u$ is constant.
Proof. Consider $w=up_1$. By assumption $w\in V$, and $w(0)=0$ by construction, so $w\in V_0$ by Claim 1. Hence $u$ is a polynomial, and we write it as $u = u(0)1+ h$ where $h\in V_0$. We need to show that $h=0$.
Consider $ua$; by assumption, $ua\in V$, and $(ua)(0)=u(0)$. So, using (the proof of) Claim 1 again, we must have $ha \in V_0$. But the set $$\{p_n \colon n\in{\mathbb N}\cup\{p_n a \colon n\in {\mathbb N}\}$$ is linearly independent: one way to see this is to note that if $g$ is any non-zero polynomial then $ga$ has at most finitely many zeros on the interval $[0,1]$. It follows that $ha=0$ and since $a(x)=e^{-x}$ this forces $h=0$ as required. $\qquad\Box$
Claim 3. $p_1a$ belongs to the closed, balanced, convex hull of $\{3p_n \colon n\in {\mathbb N}\}$.
Proof. Considering the Taylor series for $xe^{-x}$, we have $$ p_1a = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(n-1)!} p_n $$ which converges absolutely as a series in $C[0,1]$; note that $$ \sum_{n=1}^\infty \left\vert \frac{(-1)^{n-1}}{(n-1)!} \right\vert = e < 3. \qquad\qquad\Box$$
The counterexample. Let $B$ be the convex balanced hull of $\{a\}\cup\{3p_n \colon n\in{\mathbb N}\}$. Then $p_1 B \subseteq \overline{B}$ by Claim 3, and since multiplication by $p_1$ is a continuous norm-1 operator on $C[0,1]$, it follows that $p_1\overline{B}\subseteq \overline{B}$. On the other hand, if $u \in C[0,1]$ and $uB\subseteq B$ then in particular we have $up_1\in V$ and $ua\in V$, which by Claim 2 forces $u$ to be constant. Clearly $p_1$ cannot be approximated by constant functions, and so — in the notation/perspective of the original question — the multipliers of $B$ are not dense in the multipliers of $\overline{B}$.
