Let $G$ be any locally compact Hausdorff group equipped with a $\sigma$-finite measure $\mu$. Consider the faithful representation $\pi:L^\infty(G)\to B(L^2(G))$ such that $\pi(f)(g)=fg$. Define $S_1(G)=\pi^{-1}(B_1(L^2(G))$, the inverse image of trace class operators in $B(L^2(G))$. I am trying to investigate the algebra $S_1(G)$.
Q1. When is $S_1(G)$ non-empty/non-trivial? (For example if $G=\mathbb Z$ , then $S_1(G)=\ell^1(\mathbb Z)$, but if $G=S^1$ then we have $S_1(G)=\{0\}$ because $\{e^{inx}\}$ is an orthonormal basis for $L^2(S^1)$ and $$\sum_{n=1}^\infty\left<\pi(|f|)e_n,e_n\right>=\sum_{n=1}^\infty\int_\mathbb R|f|d\mu<\infty$$ if and only if $f=0$ . Also what if $G=\mathbb R$?)
Q2. Is the collection $S_1(G)$ independent of faithful representation?
Fix $f\in L^\infty (G)$. You have $\sigma(\pi(f))=\overline{f(G)}$. If $\pi(f)$ is compact its spectrum is either finite or a set whose only accumulation point is $0$. So $f$ is necessarily a step function, $f=\sum_j\alpha_k\,1_{E_j}$, where the projections $\pi(1_{E_j})$ are finite-rank except when $\alpha_j=0$.
Note that the topological struture of $G$ is irrelevant here. What matters is the measure $\mu$ (in your examples you seem to be taking the counting measure on $\mathbb Z$ and the Lebesgue measure on $S^1$ and $\mathbb R$). Write $G$ as a union $G=F\cup \bigcup_{t\in D} \{t\}$, where $F$ is diffuse (no atoms) and $D$ consists of the atoms of $\mu$. If we have $\mu(E_j\cap F)>0$ for even a single $j$, then the projection $1_{E_j}$ is not finite-rank (as we can construct an infinite chain of proper subprojections). So we may write $f=\sum_{t\in D} \beta_t 1_{\{t\}}$. And the set $\left\{\tfrac1{\mu(\{t\})^{1/2}}\,1_{\{t\}}\right\}$ is orthonormal, so $$ \operatorname{Tr}(|\pi(f)|)=\sum_{t\in D}|\beta_j|. $$ It follows that $$S_1(G)=\{f:\ \operatorname{supp}f\subset D \text{ and } \sum_{t\in D}|f(t)|<\infty\}$$
Finally, the construction does depend on the representation. Take $G=\prod_{n\in\mathbb N}\mathbb Z$ with the counting measure. Then $G\simeq \prod_{n\in\mathbb N}G$. Now map $f\longmapsto \rho(f)\in B(L^2(\prod_{n\in\mathbb N}G))$, where $\rho(\{g_n\})=\{fg_n\}$. Then $\rho(f)$ is not compact for any $f\ne0$, as each projection $\rho(1_{\{t\}})$ is now infinite.