Is the group of units of a finite ring cyclic?

A quick google search finds the paper Finite Rings Having a Cyclic Multiplicative Group of Units by Robert W. Gilmer, Jr. which settles the question as follows:

This is complemented by proving that every finite commutative ring is a direct sum of primary rings and that its group of units is the direct product of the group of units of the primary rings, and is cyclic iff each part is cyclic and they have coprime orders.

Given any ring $A$ with non-trivial group of units $A^\times$, the ring $A\times A$ will have a non-cyclic group of units.

This is because for rings $A$ and $B$ one has $(A\times B)^\times=A^\times\times B^\times$ and if $G$ is a non-trivial group the group $G\times G$ is never cyclic.

No, $\mathbb Z/8$ has unit group $\{1,3,5,7\}$ mod 8 which is a 2,2 group.

In number-theoretic situations there are coherent things that can be said, and/but in general I think nothing decisive can be said.

In the case of a field $K$, any finite subgroup of the multiplicative group is indeed cyclic. Ruffini's theorem holds only for fields (and for domains). Thus, if $A^\times$ is finite, the equation

$$x^n-1=0$$

holds for $n=|A^\times|$ and $x\in A$. However, since $(x-a)(x-b)=0$ does not imply that $x=a$ or $x=b$ anymore, it becomes complicated in the case of non-integral rings.

For $n$ a natural number, the ring ${\mathbb Z}/n{\mathbb Z}$ has a cyclic multiplicative group for $n=p^k$ or $n=2p^k$, but e.g. in the case where $n$ has at least two different odd prime divisors then it becomes non-cyclic, simply because

$$I1.\text{ }U({\mathbb Z}/p^{k}{\mathbb Z})={\mathbb Z}/p^{k-1}(p-1){\mathbb Z}$$

and by the Chinese remainder theorem you have

$$I2.\text{ }U({\mathbb Z}/n{\mathbb Z})=\oplus U({\mathbb Z}/p_i^{e_i}{\mathbb Z})=\oplus {\mathbb Z}/p_i^{e_i-1}(p-1){\mathbb Z}.$$

Clearly, the l.c.m. of the orders of the subgroup on the right strictly divides the order of the group, just because there are at least two even numbers in the list.