TITLE: What is the current status of the Kaplansky zero-divisor conjecture for group rings?
QUESTION [59 upvotes]: Let $K$ be a field and $G$ a group. The so called zero-divisor conjecture for group rings asserts that the group ring $K[G]$ is a domain if and only if $G$ is a torsion-free group.
A couple of good resources for this problem that gives some historical overview are:

Passman, Donald S. The algebraic structure of group rings. Pure and Applied Mathematics. Wiley-Interscience [John Wiley & Sons], New York-London-Sydney, 1977.
Passman, Donald S. Group rings, crossed products and Galois theory. CBMS Regional Conference Series in Mathematics, 64. Published for the Conference Board of the Mathematical Sciences, Washington, DC; by the American Mathematical Society, Providence, RI, 1986.

The conjecture has been proven affirmative, when $G$ belongs to special classes of groups. I tried to write down some of the history:

Ordered groups (A.I. Malcev 1948 and B.H. Neumann 1949)
Supersolvable groups (E. Formanek 1973)
Polycyclic-by-finite groups (K.A. Brown 1976, D.R. Farkas & R.L. Snider 1976)
Unique product groups (J.M. Cohen, 1974)

Here are my questions:

Was Irving Kaplansky the first one to state this conjecture? Can someone provide me with a reference to a paper or book that claims this?
Since the publications of Passman's expository note (above) in 1986, has there been any major developments on the problem? Are there any new classes of groups that will yield a positive answer to the conjecture? Can someone help me to extend my list above?

The zero-divisor conjecture (let's denote it by "(Z)") is related to the following two conjectures:
(I): If $G$ is torsion-free, then $K[G]$ has no non-trivial idempotents.
(U): If $G$ is torsion-free, then $K[G]$ has no non-trivial units.
Now, if $G$ is torsion-free, then one can show that:
(U) $\Rightarrow$ (Z) $\Rightarrow$ (I).
Has there been any developments, since 1986, to any partial answers on conjecture (U)? Passman claims that "this is not even known for supersolvable groups". Is this still the case?
I want to point out that this post is related to another old MO-post.

REPLY [26 votes]: Apologies for the self-promotion, but there is now a counterexample to the unit conjecture (U) with $K=\mathbb{F}_2$ and virtually abelian $G = \langle a, b \,|\, (a^2)^b=a^{-2}, (b^2)^a=b^{-2} \rangle$ (as mentioned in David Craven's answer) in arXiv:2102.11818
.