I don't know a name, but I'll give you a different characterization. Biconnectivity is sufficient but too strong, while "having minimum degree at least 2" is necessary but too weak. I think this is a necessary and sufficient condition:

$G$ has minimum degree at least 2, and if v is a cutvertex of $G$, then there is some new connected component of $G - v$ with at least two vertices adjacent to v.

This is equivalent to: Removing any vertex of degree d increases the total number of connected components by at most $d-2$. Some generalization of this property may have a name.

I don't know a name, but I'll give you a different characterization. Biconnectivity is sufficient but too strong, while "having minimum degree at least 2" is necessary but too weak. I'm almost certain this is a necessary and sufficient condition:

$G$ has minimum degree at least 2, and if v is a cutvertex of $G$, then there is some new connected component of $G - v$ with at least two vertices adjacent to v.

Here's a proof of sufficiency: If v is not a cutvertex of $G$, then pick any two vertices adjacent to v. There's a path between them not going through v (since $G - v$ is connected), so v is contained in a cycle.

If v is a cutvertex of $G$, then pick the two vertices adjacent to v that are in the same connected component of $G - v$. There's a path between them that extends to a cycle containing v.

Now, a proof of necessity. Suppose that $G$ has a cutvertex $v$ whose removal does create deg(v)-1 new connected components. Then $v$ can't lie in a cycle. (This is easy to check.)

This characterization is equivalent to: Removing any vertex of degree d increases the total number of connected components by at most $d-2$. Some generalization of this property may have a name.