$\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\R}{\mathbb{R}}
\newcommand{\F}{\mathcal{F}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}}
\newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$ 

Let us also use a comment by the OP, stating that the interval (say $I$) of density is of nonzero length and only finitely many $c_k$'s are outside of $I$. 

Let us then show that such a sequence of $c_k$'s cannot exist (and so, any bound on  $\sup_{k\leq m}\{c_{k}\}-\inf_{k\leq m}\{c_{k}\} $ will be true). 

Indeed, suppose the contrary, that there is a sequence $(c_n)_{n\ge0}$ of positive reals such that  $$a_m:=\dfrac{1}{m}\sum_{k=0}^{m-1}c_{k}\sim\prod_{k=0}^{m-1} c_{k}=:b_m$$ 
as $m\to\infty$. Since $1\in I$ and $I$ is of nonzero length, there is some $\de\in(-1,1)\setminus\{0\}$ such that $1+\de\in I$. therefore and because the $c_k$'s are dense in $I$, 
\begin{equation*}
	c_m=(1+\de)(1+o(1)) \tag{1}
\end{equation*}
infinitely often (i.o.), that is, $c_{m_j}=(1+\de)(1+o(1))$ for some increasing sequence $(m_j)$ of natural numbers as $j\to\infty$. So, 
\begin{equation*}
	b_{m+1}=b_mc_m=b_m(1+\de)(1+o(1))=a_m(1+\de)(1+o(1)) \tag{2}
\end{equation*}
i.o. On the other hand,  
\begin{equation*}
	a_{m+1}=\frac{m}{m+1}a_m+\frac{c_m}{m+1}=a_m(1+o(1))+\frac{c_m}{m+1}. \tag{3}
\end{equation*}
i.o. Since $b_{m+1}=a_{m+1}(1+o(1))$, (2) and (3) imply that i.o.
\begin{equation*}
	a_m(1+\de)(1+o(1))=a_m(1+o(1))+\frac{c_m}{m+1}, 
\end{equation*}
whence $a_m[(1+\de)(1+o(1))-(1+o(1))]=\frac{c_m}{m+1}\sim\frac{1+\de}m$ and 
\begin{equation*}
	a_m\sim\frac{1+\de}\de\,\frac1m. \tag{4}
\end{equation*}
It follows that necessarily $\de>0$, that is, $I\subseteq[1,\infty)$. 

Since only finitely many $c_k$'s are outside of $I$, we see that the positive numbers $b_m$ are nondecreasing in all large enough $m$, and so, $b_m$ goes to some $b\in[1,\infty]$ as $m\to\infty$. But this contradicts (4), since $a_m\sim b_m$.