what is the tensor product of two anti-commutative DG-algebras?

The correct setting for differential graded vectors spaces is as follows. Recall first the category of $mathbb Z$-graded vector spaces. As a category this consists of functors from the set $mathbb Z$ (thought of as a category with no morphisms) to $operatorname{Vect}$, i.e. objects consist are sequences of vector spaces, and morphisms are sequences of linear maps. (There are variations: one can insist that the vector spaces be trivial except for finitely many of them, for example, and that the non-trivial vector spaces be finite-dimensional.) As a monoidal category, the tensor structure adds degree. Only when you introduce the braiding does the category become interesting: we will take the "Koszul" braiding, so that classically odd-degree terms "anticommute". This braiding is symmetric.

Within the symmetric monoidal category $mathbb Ztext{-Vect}$ of $mathbb Z$-graded vector spaces there is a special Lie algebra, which is the unique (necessarily commutative) Lie algebra structure on the $mathbb Z$-graded vector space with one dimension in degree $1$ and all other degrees trivial. (The only bracket is the $0$ one because, by construction, the bracket must add degree, and so must land in the degree-two part, which is zero-dimensional.) Suggestively calling this Lie algebra $mathfrak{d!g}$, a differential graded vector space is nothing more nor less than a $mathfrak{d!g}$-module (in $mathbb Ztext{-Vect}$).

Let $mathfrak{d!g}text{-mod}$ denote the category of representations of $mathfrak{d!g}$. It is a symmetric monoidal category on account of it being the representation theory of a Lie algebra: the symmetric monoidal structure is inherited from $mathbb Ztext{-rep}$, so in particular there is the Koszul rule. A differential graded algebra is an algebra object in this category, and it is "anticommutative" in the classical sense if it is commutative in the categorical sense: the symmetric structure (the Koszul rule) determines for any two $mathfrak{d!g}$-modules $A,B$ a canonical isomorphism $text{flip}_{A,B}: Aotimes B to Botimes A$, and an algebra $m_A: Aotimes A to A$ is commutative if $m_A = m_A circ text{flip}_{A,A}$.

Given two algebras $(A,m_A),(B,m_B)$ in any symmetric monoidal category, their tensor product is the algebra structure on $Aotimes B$ given by: $$m_{Aotimes B} = (m_A otimes m_B) circ (text{id}_A otimes text{flip}_{A,B} otimes text{id}_B) : A otimes B otimes A otimes B to Aotimes B$$ If $A,B$ are both commutative, so is $Aotimes B$.

This categorical mumbo-jumbo exactly recovers the multiplication that you are looking for. I hope also that it illustrates that it is very naturally part of a larger story, and does not come out of the blue.