I mean what the multiplication law of the tensor product is.

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 nontrivial vector spaces be finitedimensional.) 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 odddegree terms "anticommute". This braiding is symmetric. Within the symmetric monoidal category $\mathbb Z\text{Vect}$ 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 degreetwo part, which is zerodimensional.) 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 $\mathbb Z\text{Vect}$). 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 $\mathbb Z\text{rep}$, 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 $\text{flip}_{A,B}: A\otimes B \to B\otimes A$, and an algebra $m_A: A\otimes A \to A$ 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 $A\otimes B$ given by: $$m_{A\otimes B} = (m_A \otimes m_B) \circ (\text{id}_A \otimes \text{flip}_{A,B} \otimes \text{id}_B) : A \otimes B \otimes A \otimes B \to A\otimes B$$ If $A,B$ are both commutative, so is $A\otimes B$. This categorical mumbojumbo 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. 

