Product of two elements not in a subgroup is in a subgroup?

This certainly can be true. If $H$ is a subgroup of index $2$, then it is necessarily true for any $b_1,b_2\not\in H$. (Example: in $\mathbb{Z}$, odd + odd = even ) Even if $H$ has index greater than $2$, it can be true. You only need that $b_1$ and $b_2$ belong to cosets of $H$ that are inverses in the quotient group.

Sure. Take $G=(\mathbb{Q},+)$ and $H=\mathbb Z$. Then $\frac12+\frac12=1\in\mathbb{Z}$, but $\frac12\notin\mathbb Z$.

We can say more. Given $H$ a proper subgroup of $G$, and an element $b_1 \in G$ that is not in $H$, there are $|H|$ choices for $b_2$, none of them in $H$, such that $b_1b_2 \in H$. Note that $b_1^{-1} \not \in H$ and $b_1b_1^{-1}=e \in H$. Then for each $h \in H$ we have $b_1^{-1}h \not \in H$ and $b_1(b_1^{-1}h)\in H$