Let $C$ and $D$ be $B$ algebras, where $B$ is a commutative ring with unity. And consider $C \otimes_B D$?
I first thought $1 \otimes 1$ was always the unity element in tensor products, but then I saw examples where $1 \otimes 1$ is in fact $0$. I thought perhaps the following is true?: $1 \otimes 1$ is the unity element in $C \otimes_B D$ if and only if $C \otimes_B D \not = \{ 0 \}$
Yes, $1\otimes1$ is the unity in the tensor product algebra, because it satisfies $$ (1\otimes1)(c\otimes d)=c\otimes d=(c\otimes d)(1\otimes1) $$ so the property also transfers to every element of $C\otimes_BD$.
If it happens that $1\otimes1=0$, then you can conclude that $C\otimes_BD=\{0\}$.
The ring $\{0\}$ as a unity, namely $0$. There is no requirement that in a ring the unity (identity) is nonzero.
The condition $1\ne0$ is imposed to domains (and so to fields), because there are very good reasons not to consider $\{0\}$ a domain.