Let $G$ be a locally compact group, $C_0(G)$ the $C^*$-algebra of continuous functions on $G$ that vanish at infinity, $C_b(G)$ the $C^*$-algebra of bounded continuous functions on $G$. We know that $C_b(G)$ is the multiplier algebra of $C_0(G)$, and we denote the strict topology on $C_b(G) = \mathcal{M}\bigl( C_0(G) \bigr)$ by $\beta$.
Now for a strongly continuous unitary representation $(\pi, H)$ of $G$, functions of the form $\omega_{\pi,\eta,\xi} : g \in G \to (\pi(g)\eta \mid \xi) \in \mathbb{C}$ are in $C_b(G)$, and we call them matrix coefficients of the representation $\pi$. Since we can form direct sum and tensor product of two strongly continuous unitary representations, as well as the contragredient representation, we see that (the linear span of) matrix coefficients of all strongly continuous unitary representations of $G$ form a $*$-subalgebra $A_0(G)$ of $C_b(G)$.
Question. Is $A_0(G)$ strictly dense in $C_b(G)$, i.e. with respect to the $\beta$-topology?
Note that in the compact case, $C_b(G) = C_0(G) = C(G)$ and the $\beta$-topology is the same as the norm topology on the $C^*$-algebra $C(G)$ of continuous functions on $G$. In this case, The answer to the question is affirmative by Peter-Weyl. In the case where $G$ is discrete, then one can check easily that all finitely supported functions on $G$ are already matrix coefficients of the left (or right) regular representation, so the answer to the question is again affirmative. Based on these considerations, here are some sub-questions with some bias on their possible answers.
Q1. Does the question have an affirmative answer for unimodular $G$?
Q2. Can we construct some counter-example for non-unimodular $G$?
Q3. Does the question have an affirmative answer if $G$ is a real Lie group? What if the Lie group $G$ is nilpotent, or solvable, or semisimple/reductive?
Best Answer
First, some remarks that may help with literature-searching.$\newcommand{\fsnorm}[1]{{\Vert#1\Vert}_{\rm B}}$ $\newcommand{\supnorm}[1]{{\Vert#1\Vert}_\infty}$
The algebra you have denoted by $A_0(G)$ is known as the Fourier--Stieltjes algebra of $G$, and is usually denoted by $B(G)$, so I will do that from now on. $B(G)$ has been much studied: it turns out to be complete in a natural submultiplicative norm $\fsnorm{\cdot}$, so it is a commutative Banach algebra of functions on $G$; in fact it is also a dual Banach space for this norm, and multiplication is separately weak-star continuous, so it is an example of a dual Banach algebra.
This norm dominates the supremum norm, so $\fsnorm{\cdot}$-convergence implies $\supnorm{\cdot}$-convergence. If one takes the $\fsnorm{\cdot}$-closure inside $B(G)$ of $B(G)\cap C_c(G)$, the resulting algebra is an ideal in $B(G)$ called the Fourier algebra of $G$, usually denoted by $A(G)$. One can show that $A(G)$ is a $\supnorm{\cdot}$-dense subalgebra of $C_0(G)$ that is closed under conjugation and separates points of $G$.
(Aside: in general $B(G)\cap C_0(G)$ is larger than $A(G)$.)
Given some $f\in C_b(G)$, an $\varepsilon>0$, and a finite set $E\subset C_0(G)$, it suffices to find $h\in B(G)$ such that $\supnorm{fg-hg}\leq \varepsilon$ for all $g\in E$.
Since $C_0(G)$ is $\beta$-dense in $C_b(G)$ we can find $k \in C_0(G)$ such that $\supnorm{ fg-kg } \leq \varepsilon/2$ for all $g\in E$. Since $B(G)\cap C_c(G)$ is $\supnorm{\cdot}$-dense in $C_0(G)$ we can find $h\in B(G)\cap C_c(G)$ such that $\supnorm{k-h} \max_{g\in E} \supnorm{g} \leq\varepsilon/2$. Then $$ \supnorm{ fg -hg} \leq \supnorm{fg-kg} + \supnorm{k-h}\supnorm{g} \leq \varepsilon \quad\hbox{for all $g\in E$,} $$ as required.