A braiding gives the tensor product algebra structure

The proposed multiplication first moves the middle $B\otimes A$ pair to $A\otimes B$ by $c_{B,A}$ and then applies $m_A$ and $m_B$. With a unit input, naturality of the braiding with the unitors reduces the crossing to the identity, so the left and right unit composites are $m_A({\eta }_A\otimes 1_A)=1_A$ and $m_B({\eta }_B\otimes 1_B)=1_B$, or their right-unit counterparts. Thus ${\eta }_A\otimes {\eta }_B$ is a unit.

On three tensor factors $A\otimes B\otimes A\otimes B\otimes A\otimes B$, the left-associated product first crosses $B_1$ past $A_2$, then expands the crossing of $B_1\otimes B_2$ past $A_3$ into the crossings of $B_2$ and $B_1$ past $A_3$. The right-associated product first crosses $B_2$ past $A_3$, then expands the crossing of $B_1$ past $A_2\otimes A_3$ into the crossings past $A_2$ and $A_3$. The two expansions use the two hexagons [F1]. The crossing of $B_1$ past $A_2$ and that of $B_2$ past $A_3$ involve disjoint tensor factors and commute by functoriality, so both composites perform the same three crossings in the same resulting order. They then apply respectively $m_A(m_A\otimes 1_A)$ and $m_B(m_B\otimes 1_B)$, or $m_A(1_A\otimes m_A)$ and $m_B(1_B\otimes m_B)$; these agree by the algebra associativity equations in [F2]. Hence $m_{A\otimes B}$ is associative.