I can't see that your map $\phi:n\
ewcommand{\ot}{\otimes}\ot b\mapsto b\ot n$ is a $B\ot_A B$-module homomorphism. $$\phi((b_1\ot b_2)(n\ot b))=\phi(b_1n\ot b_2b)= b_2b\ot b_1 n$$ but $$(b_1\ot b_2)\phi(n\ot b)=(b_1\ot b_2)(b\ot n)=b_1b\ot b_2n.$$ Those look different to me.