From the picture above, we see that the union of the half-disks will be a quarter-circle with radius $2$, and therefore area $\pi$. To prove that this is the case, we first prove that the boundary of every half-disk intersects the quarter-circle with radius $2$, and then that the half-disk is internally tangent to the quarter-circle at that point. This is sufficient because it is clear from the diagram that we need not worry about covering the interior of the quarter-circle.
Let $O$ be the origin. For a given half-disk $D_{y}$, label the vertex on the $y$-axis $A$ and the vertex on the $x$-axis $B$. Let $M$ be the midpoint of line segment $\overline{A B}$. Draw segment $O M$, and extend it until it intersects the curved boundary of $D_{y}$. Label the intersection point $C$. This construction is shown in the diagram below.

We first prove that $C$ lies on the quarter-circle, centered at the origin, with radius $2$. Since $M$ is the midpoint of $\overline{A B}$, and $A$ is on the $y$-axis, $M$ is horizontally halfway between $B$ and the $y$-axis. Since $O$ and $B$ are on the $x$-axis (which is perpendicular to the $y$-axis), segments $\overline{O M}$ and $\overline{M B}$ have the same length. Since $M$ is the midpoint of $\overline{A B}$, and $A B=2$, $O M=1$. Since $D_{y}$ is a half-disk with radius $1$, all points on its curved boundary are $1$ away from its center, $M$. Then $C$ is $2$ away from the origin, and the quarter-circle consists of all points which are $2$ away from the origin. Thus, $C$ is an intersection of the half-disk $D_{y}$ with the positive quarter-circle of radius $2$.
It remains to show that the half-disk $D_{y}$ is internally tangent to the quarter-circle. Since $\overline{O C}$ is a radius of the quarter-circle, it is perpendicular to the tangent of the quarter-circle at $C$. Since $\overline{M C}$ is a radius of the half-disk, it is perpendicular to the tangent of the half-disk at $C$. Then the tangent lines of the half-disk and the quarter-circle coincide, and the half-disk is tangent to the quarter-circle. It is obvious from the diagram that the half-disk lies at least partially inside of the quarter-circle, the half-disk $D_{y}$ is internally tangent to the quarter-circle.
Then the union of the half-disks is a quarter-circle with radius $2$, and has area $\pi$.

Solution:
Answer: $\pi$

From the picture above, we see that the union of the half-disks will be a quarter-circle with radius $2$, and therefore area $\pi$. To prove that this is the case, we first prove that the boundary of every half-disk intersects the quarter-circle with radius $2$, and then that the half-disk is internally tangent to the quarter-circle at that point. This is sufficient because it is clear from the diagram that we need not worry about covering the interior of the quarter-circle.
Let $O$ be the origin. For a given half-disk $D_{y}$, label the vertex on the $y$-axis $A$ and the vertex on the $x$-axis $B$. Let $M$ be the midpoint of line segment $\overline{A B}$. Draw segment $O M$, and extend it until it intersects the curved boundary of $D_{y}$. Label the intersection point $C$. This construction is shown in the diagram below.

We first prove that $C$ lies on the quarter-circle, centered at the origin, with radius $2$. Since $M$ is the midpoint of $\overline{A B}$, and $A$ is on the $y$-axis, $M$ is horizontally halfway between $B$ and the $y$-axis. Since $O$ and $B$ are on the $x$-axis (which is perpendicular to the $y$-axis), segments $\overline{O M}$ and $\overline{M B}$ have the same length. Since $M$ is the midpoint of $\overline{A B}$, and $A B=2$, $O M=1$. Since $D_{y}$ is a half-disk with radius $1$, all points on its curved boundary are $1$ away from its center, $M$. Then $C$ is $2$ away from the origin, and the quarter-circle consists of all points which are $2$ away from the origin. Thus, $C$ is an intersection of the half-disk $D_{y}$ with the positive quarter-circle of radius $2$.
It remains to show that the half-disk $D_{y}$ is internally tangent to the quarter-circle. Since $\overline{O C}$ is a radius of the quarter-circle, it is perpendicular to the tangent of the quarter-circle at $C$. Since $\overline{M C}$ is a radius of the half-disk, it is perpendicular to the tangent of the half-disk at $C$. Then the tangents lines of the half-disk and the quarter-circle coincide, and the half-disk is tangent to the quarter-circle. It is obvious from the diagram that the half-disk lies at least partially inside of the quarter-circle, the half-disk $D_{y}$ is internally tangent to the quarter-circle.
Then the union of the half-disks is be a quarter-circle with radius $2$, and has area $\pi$.