Ring theory: well-definedness of associates without commutativity Definition of associates:

An element $u$ of a commutative ring with unity $R$ is a unit of $R$ if $u$ divides $1$, that is, if $u$ has a multiplicative inverse in $R$. Two elements $a,b \in R$ are associates in $R$ if $a=bu$, where $u$ is a unit in $R$.

In a non-commutative ring with unity, if $a=bu$ for some invertible $u$, must there be invertible $v$ such that $a=vb$?

No.  For instance, let $k$ be a field and $R=M_2(k)$, and take $a=\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$ and let $b=\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$.  Then $a=bu$, where $u=\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$ is invertible, but $a$ cannot be $vb$ for any $v$ since the kernel of $b$ is not contained in the kernel of $a$.
