Rings $R$ such that every [regular] square matrix with entries in $R$ is equivalent to an upper triangular matrix

As Luc Guyot mentioned, check out Kaplansky's paper Elementary Divisors and Modules from 1949.

Kaplansky calls a ring Hermite when every $1 \times 2$ matrix is equivalent to a diagonal matrix, and shows that equivalently a ring is Hermite iff for every matrix $M$ there exists an invertible matrix $U$ such that $MU$ is upper triangular.

To get a finite product decomposition result like Yohe's, we obviously need to assume that $R$ has finitely many minimal primes.

On the other hand, we have the following

Theorem If $R$ is a Bézout ring with finitely many minimal primes then there is a finite set of idempotents $e_i$ such that the ring $e_iR$ is a Hermite ring with a unique minimal prime and $R \cong \prod e_iR$. Hence $R$ is Hermite.

This is Theorem 2.2 in Elementary Divisor Rings and Finitely Presented Modules by Larsen, Lewis, and Shores.

Maybe we can say more about the structure of these summands. For example, it is easy to show that in any Bézout ring with a unique minimal prime $P$, the ideal $P$ is essential unless $R$ is a domain. Indeed, if $I \cap P = 0$, then $\operatorname{Ann}(a) = P$ for any $a \in I$ and clearly any $a \in I$ is nonzero in every localization of $R$. Since $R$ locally has totally ordered ideals, this implies that $P$ is locally $0$, hence $0$, i.e. $R$ is a domain.

So to sum up

Conclusion Let $R$ be a ring with finitely many minimal prime ideals. Then the following are equivalent:

$\ \ (1)$ $R$ is a finite direct product of Bézout rings each of which is either a domain or has a unique minimal prime ideal which is essential.

$\ \ (2)$ Matrices over $R$ are equivalent to triangular matrices.

I'm not sure if you can say very much in general about the structure of Bézout rings with essential unique minimal prime ideal. But, here's one observation: A Bézout ring with a unique minimal prime ideal $P$ has the property that every non-nilpotent element divides every nilpotent element.

Indeed, let $b$ be nilpotent and let $a$ be not nilpotent, i.e. $b \in P$ and $a \notin P$. By Bézoutness, pick $c,d,u,v,r$ such that $ac + bd = r$, $ru = a, rv = b$. Deduce that $r \notin P, v \in P$. It follows that $cv + du -1 \in P$, and we deduce that $du$ is a unit. Therefore $a \mid b$.

Note that if $R$ is additionally Noetherian, we can easily deduce that $P$ is the unique prime ideal of $R$, which exactly recovers Yohe's result. To see this, note first that it suffices to assume $R$ is local with maximal ideal $M$, in which case it has totally ordered ideals and $\bigcap_n M^n = 0$, which implies $M^n \subseteq P$ for some $n$ and thus $M \subseteq P$. Without the Noetherian hypothesis, $R$ can have infinite Krull dimension.

As for your 3rd question, I don't have much to say off the top of my head except that it seems really hard. From your own observations, any ring which is its own total ring of fractions (i.e. regular elements are units) will have this property, and that is quite a broad class of rings, members of which sometimes have ostensibly very little in common.

I will answer only the second and third question in your list, the first is too general and open ended.

Take $R=K[x,y,z,w]$, polynomial ring in four variables over a field $K$. Take the $2\times 2$ matrix $M$ consisting of the four variables as entries. It is regular, but it is not equivalent to an upper triangular matrix.

If it is, then by determinant considerations, one of the diagonal entries have to be a non-zero constant since $\det M$ is irreducible in $R$. We have $uMv=N$ with $u,v$ invertible, $N$ has a non-zero constant entry. But, put all variables equal to zero and then $M(0)=0$ and then $N(0)$ must be zero, which is impossible, since one entry is a non-zero constant.