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

Now write $b = \sum_i a_ib_i$. Then
  $$b\otimes 1 - 1 \otimes b = \sum_i (a_ib_i \otimes 1 - 1 \otimes a_ib_i)=\sum_i (a_ib_i\otimes 1 - a_i \otimes b_i)=\sum_i(a_i \otimes 1)(b_i\otimes 1 - 1 \otimes b_i)$$ Hence $I=(b_i\otimes 1 - 1 \otimes b_i\mid i=1,\ldots,n)$ as desired.

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.