Product of classes of associates in a commutative ring/rng

Question

Let $[a]$ be the equivalence class of associates of an element $a$ of a ring/rng.

Two elements of a ring/rng are associates ($\sim$) if they generate the same principal ideal.

$[a] \cdot [b] \sim [a \cdot b]$ in any commutative rng $R$ in the following way:

1. For any element $a'$ from $[a]$ and $b'$ from $[b]$: $a' \cdot b' \sim c$ for some $c$ from $[a \cdot b]$;
2. For any element $c$ from $[a \cdot b]$: $c \sim a' \cdot b'$, for some $a'$ from $[a]$ and $b'$ from $[b]$.

Proof:

1. If $a' \sim a$, then $a' = r \cdot a + na$ for some $r$ from $R$ and an integer $n$;
   If $b' \sim b$, then $b' = s \cdot b + mb$ for some $s$ from $R$ and an integer $m$; $a' \cdot b' = (r \cdot a + na) \cdot (s \cdot b + mb) = (r \cdot s + ns + mr) \cdot (a \cdot b) + nm(a \cdot b)$;
   Thus, $a' \cdot b'$ is in the principal ideal of $a \cdot b$;
   Exchanging $a'$ with $a$, and $b'$ with $b$ in the formulas above:
   $a \cdot b$ is in the principal ideal of $a' \cdot b'$;
   Therefore, $a' \cdot b' \sim a \cdot b$, where $a \cdot b$ is an element of $[a \cdot b]$.

2. $c \sim a \cdot b$ for any element $c$ from $[a \cdot b]$ by the definition of the class of associates.

I am wondering in which types of rings/rngs we can write $[a] \cdot [b] = [a \cdot b]$ meaning:

1. For any element $a'$ from $[a]$ and $b'$ from $[b]$: $a' \cdot b' = c$ for some $c$ from $[a \cdot b]$;
2. For any element $c$ from $[a \cdot b]$: $c = a' \cdot b'$, for some $a'$ from $[a]$ and $b'$ from $[b]$.

The first part of the definition is identical to the one for $[a] \cdot [b] \sim [a \cdot b]$
since $a' \cdot b' \sim a \cdot b$ means $a' \cdot b'$ is an element of $[a \cdot b]$.

I need help with the second part of the definition.
The widest class of rings where $c \sim a \cdot b$ means $c = a' \cdot b'$ for some $a' \sim a$ and $b' \sim b$
I found so far is integral domains:

If $c \sim a \cdot b$ in an integral domain, then $c = u \cdot (a \cdot b)$ for some unit $u$;
Then $c = (u \cdot a) \cdot b$, where $u \cdot a \sim a$ and $b \sim b$.

However, it looks like the formula $[a] \cdot [b] = [a \cdot b]$ works for all cyclic rings/rngs.
Is it possible to find a wider class than integral domains?
Is it true in commutative principal ideal rings/rngs?

Update

The formula $[a] \cdot [b] = [a \cdot b]$ works for all cyclic rngs:

1. $[a] \cdot [b] = [a \cdot b]$ in a prime-power cyclic rng
   (since $a \sim p^k$ for any element $a$ of a prime-power cyclic rng
   https://math.stackexchange.com/a/3513734/427611);

2. $[(a, b)] = ([a], [b])$ for the direct product of rngs
   ($(a, b) \sim (c, d) \iff a \sim c \land b \sim d$);

3. Any finite cyclic rng is the direct product of prime-power cyclic rngs.

I am trying to apply the same logic to all commutative principal ideal rngs
using the Zariski–Samuel theorem (Structure theory for commutative PIR's):
$[a] \cdot [b] = [a \cdot b]$ in principal ideal rngs if
$[a] \cdot [b] = [a \cdot b]$ in special principal rngs.

I need help checking the formula for special principal rngs.