It's a (pretty easy) standard exercise in algebra to show $I=(b\otimes 1 - 1 \otimes b\mid b \in B)$. 

Let $a\in A, b,c \in B$. Then 
$$ab\otimes 1 - 1 \otimes ab = (a\otimes 1)(b\otimes 1 - 1\otimes b)$$
$$bc\otimes 1 - 1 \otimes bc = (b\otimes 1)(c\otimes 1 - 1 \otimes c) + (b \otimes 1 - 1 \otimes b)(1 \otimes c)$$
Hence if $B$ is generated as $A$-algebra by $b_1,\ldots, b_n$, then $I$ is generated as ideal in $B\otimes_AB$ by $b_i \otimes 1 - 1 \otimes b_i\,\,(i=1,\ldots,n)$ as desired.