TITLE: Explict form of $E_\infty$-morphisms between differential graded commutative algebras
QUESTION [9 upvotes]: This is a partial duplicate to this MO question, I apologize for that. I'm asking since the answers there still do not allow me to work out an answer to my question, which is a bit more specific.
Informally speaking, an $E_\infty$-algebra is an algebra which is associative and commutative up to coherent homotopies, a notion that can be properly formalized via the language of operads. However, in characteristic zero, one can can obtain an equivalent category to $E_\infty$ algebras by taking commutative differential graded algebras and inverting the quasi-isomorphisms. This is the same mechanism by which one can define the category of $L_\infty$ algebras in characteristic zero by starting with differential graded Lie algebras and inverting quasi-isomorphisms.
However, in the $L_\infty$-case it is well known that $L_\infty$-morphisms between differential graded Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$ can be also given an "explicit" definition as sequences of multilinear maps $\varphi_k\colon \wedge^k \mathfrak{g}\to \mathfrak{h}[1-k]$ which satisfy a set of quadratic equations. This explicit definition obscures the theory behind it but has nice "practical" consequences. For instance one immediately sees that an $L_\infty$-morphism between two Lie algebras (seen as differential graded Lie algebras concentrated in degree zero) is precisely the same thing as a Lie algebra morphism between them, or that an $L_\infty$-morphism from a Lie algebra $\mathfrak{g}$ to a differential graded Lie algebra $\mathfrak{h}$ concentrated in nonegative degree is the same thing as a morphism of differential graded Lie algebras from $\mathfrak{g}$ to $\mathfrak{h}$. Or even, if $\mathfrak{g}$ is a Lie algebra and $\mathfrak{h}=(\mathfrak{h}^{-1}\to \mathfrak{h}^0)$ is a differential graded Lie algebra concentrated in degrees -1 and 0, one has a complete, simple and explicit control on what an $L_\infty$ morphism from $\mathfrak{g}$ to $\mathfrak{h}$ would be.
I'm wondering whether something similar can be said for $E_\infty$-morphisms between differential garded commutative algebras.
For instance: can it be given an explicit description of the $E_\infty$-morphisms between a commutative algebra $A$ (seen as a differential graded commutative algebra concentrated in degree zero) and the differential graded commutative algebra $\mathbb{K}[1]\oplus \mathbb{K}$, where $\mathbb{K}$ is the base field? (the differential on $\mathbb{K}[1]\oplus \mathbb{K}$ is the zero differential and the multiplication $(\mathbb{K}[1]\oplus \mathbb{K})^0\otimes (\mathbb{K}[1]\oplus \mathbb{K})^{-1}\to (\mathbb{K}[1]\oplus \mathbb{K})^{-1}$ is the multiplication in $\mathbb{K}$).
I guess such an $E_\infty$-moprhism should reduce to the datum of two $\mathbb{K}$-linear maps $\varphi_0\colon A\to \mathbb{K}$ and $\varphi_1\colon A\otimes A\to \mathbb{K}$ satisfying suitable equations which shuld precisely encode the fact that $\varphi_0\colon A\to \mathbb{K}$ is a morphism of commutative algebras up to a homotopy given by $\varphi_1$. But I'm not able to find a rigorous statement along these lines anywhere in the literature.

REPLY [4 votes]: For future readers' convenience, I'm summarizing the comments to the original question in the form of an answer. Many thanks to Adeel, Sean Tilson, Akhil Mathew and Fernando Muro for their inputs. I'm obviously the only one to blame for eventual inaccuracies I'll be including in this answer. So, the answer is as follows:
Let $A$ and $B$ two differential graded commutative $\mathbb{K}$-algebras, where $\mathbb{K}$ is a characteristic zero field. An $E_\infty$-morphism $\varphi\colon A\to B$ is precisely a morphism of $C_\infty$-algebras between $A$ and $B$. Equivalently, it is a morphism of $A_\infty$-algebras form $A$ to $B$ such that all of its Taylor coefficients $\varphi_k\colon A^{\otimes k}\to B[1-k]$ vanish on the shuffle products.