UFD, prime and Irreducible

Solution 1:

First, let's consider the simpler case when $R$ is an integral domain. Part 2 of U1 is redundant since factorizations into primes are unique (up to associates and order), by the same simple proof as in $\,\Bbb Z$ (recall prime $\Rightarrow$ irreducible). Here U1 and U2 are both are equivalent to $R$ is a UFD. This follows by the standard proofs given in most any textbook on abstract algebra.

Factorization theory is more complicated in non-domains: basic notions such as associate and irreducible bifurcate into a few inequivalent notions. For a survey in the commutative case, see D.D. Anderson, and S. Valdes-Leon, $ $ Factorization in Commutative Rings with Zero-divisors. Below I have appended some of their results pertaining to your query (whose answer depends on the precise notion of "associate" and "irreducible" that is employed).

For noncommutative UFDs it is more convenient to employ refinement-based views of unique factorization, e.g. see Paul Cohn's 1973 Monthly survey Unique factorization domains. and see also his 1963 TAMS paper Noncommutative unique factorization domains.