Associates in Integral Domain
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Your reasoning for $(\Rightarrow)$ appears to be correct. Note that the reason you can cancel: $y = yts \implies 1 = ts$ is because we're in an integral domain.
For $(\Leftarrow)$, the reasoning would be something like this: Since $x = yd$ where $d$ is a unit, then $y \mid x$, we can rewrite $x = yd$ as $y = xd^{-1}$ since $d$ is a unit and hence $x \mid y$. Since $x \mid y$ and $y \mid x$, then $x$ and $y$ are associates.