$R$ - commutative Ring, and $a$ non-zero-divisor
$R^x$ - multiplicative Monoid? (it's called Einheitsgruppe in German, maybe unitary group?)
I started by stating, that if $a\mid b$ and $b\mid a \rightarrow a=b$.
So for $a\mid b$, $\exists x\in R$ with $xa=b$ and for $b\mid a,\exists y\in R$ with $yb=a$
Now: $b=xa=xyb=wb$, with $w=xy$.
$\stackrel{a=b}{\rightarrow} b=wa$
Now the problem that I think I have with my proof, is that R is not a given Ring. For example $a=b$ wouldn't work in $\mathbb{Z}$. Is there any other way to do the proof?
I assume you are working in a commutative ring with identity.
The problem is to show that if two elements of a ring divide each other, then they are associates, i.e. the same up to a unit multiple.
Actually this is not true in general. For example, in the ring $\mathbb{Z}[x,y,z]/(xy -z, zy - x)$ we have that $x \mid z$ and $z \mid x$, but $y$ is not a unit. Note that $x$ and $z$ are both zero divisors, as $z(y^2 - 1) = x(y^2-1) = 0$.
In general, $a \mid b$ means there exists $u \in R$ with $au = b$.
$b \mid a$ means there exists $v \in R$ with $bv = a$.
and putting these equations together...
$$buv = b$$ $$auv = a$$
It follows that if either $a$ or $b$ is not a zero divisor, then $uv = 1$ and $u,v$ are units, therefore $a$ and $b$ are associates.