Is the "set" of all sets which contain $1$ a set under ZF?
It does not contain itself, as it contains sets which contain $1$, but does not contain $1$ itself (as $1$ is not a set which contains $1$), and this has been the cause of a "failure to be a set" in all the cases that I've considered.
What axiom(s) of ZF is violated, if any, and why?
On
This class (call it "$C$") is not a set. There are a couple different ways to see this. For example:
Consider the subclass $O$ of ordinals in $C$. If $C$ is a set, so is $O$ (by Separation), but $O\cup\{0, 1\}$ is the class of all ordinals, which is not a set.
Let $D=\{C,1\}$. We have $C\in D\in C$, and this contradicts Regularity. (Specifically, the set $\{C, D\}$ has no $\in$-minimal element.)
The transitive closure of $C$ is the class of all sets, which of course isn't a set.
Note that there are really two reasons why $C$ is not a set: it's "too big," and it leads to a failure of regularity (even though as you observe it doesn't contain itself - the failure of regularity here is not as direct, but it's still present).
Such a class is as large as the class of all sets (and hence is not a set), because for each set $S$ it contains $S\cup\{ 1\}$, and hence it can be bijected with the sets $\not\owns 1$. If they lived in a set, and our original class were a set too, we would have identified two sets whose union is the class of all sets.