Let the four incenters be $I_{1}$, $I_{2}$, $I_{3}$, and $I_{4}$ with inradii $r_{1}$, $r_{2}$, $r_{3}$, and $r_{4}$ respectively (in the order given in the question). Without loss of generality, let $I_{1}$ be closer to $AB$ than $I_{2}$. Let the acute angle between $I_{1}I_{2}$ and $AB$ (and hence also the angle between $I_{3}I_{4}$ and $CD$) be $\theta$. Then
$$r_{2}-r_{1}=I_{1}I_{2} \sin \theta=I_{3}I_{4} \sin \theta=r_{4}-r_{3},$$
which implies $r_{1}+r_{4}=r_{2}+r_{3}$. Similar arguments show that $r_{1}+r_{2}=r_{3}+r_{4}$. Thus we obtain $r_{1}=r_{3}$ and $r_{2}=r_{4}$.

Now let's consider the possible positions of $W$, $X$, $Y$, $Z$. Suppose $AZ \neq CX$. Without loss of generality assume $AZ>CX$. Since the incircles of $AWZ$ and $CYX$ are symmetric about the centre of the parallelogram $ABCD$, this implies $CY>AW$. Using similar arguments, we have
$$CY>AW \Longrightarrow BW>DY \Longrightarrow DZ>BX \Longrightarrow CX>AZ,$$
which is a contradiction. Therefore $AZ=CX \Longrightarrow AW=CY$ and $WXYZ$ is a parallelogram.