When is Thom isomorphism a ring map?

Question

Let $R$ be an $E_{\infty}$-ring spectrum and $B$ be an $E_\infty$-space. Suppose we have an $E_\infty$-map $$ f: B \to BGL_1(S^0)$$ such that the composite $$f_R: B \to BGL_1(S^0) \to BGL_1(R) $$ is null, then a choice of null-homotopy produces Thom isomorphism which is a weak-equivalence $$u: Mf \wedge R \simeq B_+ \wedge R,$$

where $Mf$ is the Thom spectrum associated to $f$.

Note that, both sides of the Thom isomorphism are ring spectra.

Q: I wonder when $u$ is a ring-map?

A possible guess is that if $f_R$ is homotoped to null via infinite-loop space maps, then maybe $u$ is a ring map. I am not quite sure if this is true or how to see this.

Answer 1

One comment is that you have to be careful, because to be an $E_{\infty}$-map is not a property but rather additional structure.

You are exactly right in that what is needed is a null-homotopy of $B \rightarrow BGL_{1}(R)$ as a map of $E_{\infty}$-spaces. This is Proposition 3.16 in Omar and Toby's "A simple universal property of Thom ring spectra" when you set $A = R$.