For any spectrum $E$ we can define "spectrum with coefficients in $G$" by $EG := E \wedge MG$, where $MG$ is the Moore spectrum of type $G$.
If $E$ is an $A_{\infty}$ (resp. $E_{\infty})$ ring spectrum, will $EG$ also be $A_{\infty}$ (resp. $E_{\infty})$? If not in general, are there any specific conditions under which that might be true (e.g. $G$ finite etc)?
Thank you for any references!
This is not a full answer, but rather just an example.
Let $MC_2$ be the mod 2 Moore spectrum. Notably, it is not a ring spectrum, so even though $E = \mathbb{S}$ is $E_\infty$, the spectrum $\mathbb{S}C_2 \simeq MC_2$ is not a ring spectrum, let alone $A_\infty$ or $E_\infty$.
This shows that in general even if $E$ is a structured ring, $EG$ need not be. It also shows that the finiteness of $G$ is not the condition you are looking for.
Added later: In fact, you can just ask whether Moore spectra are highly structured. It turns out that it is generally not the case. For example, they are never $E_\infty$, and that for many $G$ they are known to not be $A_\infty$. See the answers on this mathoverflow question for details.