I'm reading Keller's Introduction to A-infinity algebras and modules to learn about $A_\infty$-algebras. For reference, an $A_\infty$-algebra $A$ is a graded $k$-vector space $A = \bigoplus_{i\in\mathbb{Z}}A^i$ equipped with linear maps $m_n:A^{\otimes n}\to A$ each of degree $2-n$ satisfying the Stasheff identities
$$\sum_{r+s+t = n} (-1)^{r+st}m_{r+1+t}(\boldsymbol{1}^{\otimes r}\otimes m_s \otimes \boldsymbol{1}^{\otimes t}) = 0$$
The map $m_1$ is a differential on $A$ and $m_2$ is a product on $A$ satisfying the graded Leibniz identity, making it a (not necessarily associative) differential graded algebra.
When describing the Stasheff identities for low values of $n$, it's written in the document I linked above (and other places I've read defining $A_\infty$-algebras) that for $n=3$ the identity
$$m_2(\boldsymbol{1}\otimes m_2 - m_2\otimes \boldsymbol{1}) = m_1m_3 + m_3(m_1 \otimes\boldsymbol{1}\otimes\boldsymbol{1}+\boldsymbol{1}\otimes m_1\otimes \boldsymbol{1} + \boldsymbol{1}\otimes \boldsymbol{1}\otimes m_1)$$
and the fact that
$$ m_1m_3 + m_3(m_1 \otimes\boldsymbol{1}\otimes\boldsymbol{1}+\boldsymbol{1}\otimes m_1\otimes \boldsymbol{1} + \boldsymbol{1}\otimes \boldsymbol{1}\otimes m_1) = \delta(m_3)$$
where $\delta$ is the differential of $\mbox{Hom}(A^{\otimes 3}, A)$, implies that $m_2$ is "associative up to homotopy". I've been trying to find a precise description of what this actually means but I've not been able to find anything. My assumption is that it means $m_2$ is homotopic to an associative product, but that somehow doesn't feel completely correct.
I realise that $m_2(\boldsymbol{1}\otimes m_2 - m_2\otimes \boldsymbol{1})$ is the associator of $m_2$, so that when it's equal to $0$ then this means that $m_2$ is an associative product, but I'm not sure how this coupled with the fact that $m_1m_3 + m_3(m_1 \otimes\boldsymbol{1}\otimes\boldsymbol{1}+\boldsymbol{1}\otimes m_1\otimes \boldsymbol{1} + \boldsymbol{1}\otimes \boldsymbol{1}\otimes m_1)$ is a coboundary has anything to do with homotopies.
If someone could help me understand I would really appreciate it!
You asked about the algebraic point of view. If $(C, d_C)$ and $(D, d_D)$ are chain complexes with differentials of degree 1 and $f, g: C \to D$ are chain maps, then they are chain homotopic, or just homotopic, if there is a sequence of maps $H: C_n \to D_{n-1}$ such that $$ f-g = d_D H + H d_C. $$ This is an equivalence relation, and one consequence is that $f$ and $g$ induce the same map on homology.
That's the situation here: one chain complex is $(A, m_1)$ and the other is $(A \otimes A \otimes A, m_1 \otimes 1 \otimes 1 + 1 \otimes m_1 \otimes 1 + 1 \otimes 1 \otimes m_1)$, and the two chain maps $m_2(m_2 \otimes 1)$ and $m_2(1 \otimes m_2)$. The Stasheff identity for $n=2$ says that $m_2$ induces a multiplication on the homology of $A$ with respect to $m_1$, and the $n=3$ identity implies that at the level of homology, the multiplication is associative. (Being an $A_\infty$-algebra is much stronger than just being associative at the level of homology, though.)