The minimum is 4, achieved by the square with vertices $(\pm 1, \pm 1)$.

\textbf{First solution:}
To prove that 4 is a lower bound, let $S$ be a convex set of the desired form. Choose $A,B,C,D \in S$ lying on the branches of the two hyperbolas, with $A$ in the upper right quadrant, $B$ in the upper left, $C$ in the lower left, $D$ in the lower right.
Then the area of the quadrilateral $ABCD$ is a lower bound for the area of $S$.

Write $A = (a,1/a)$, $B = (-b,1/b)$, $C = (-c,-1/c)$, $D = (d, -1/d)$ with $a,b,c,d > 0$.
Then the area of the quadrilateral $ABCD$ is
$$\frac{1}{2}(a/b + b/c + c/d + d/a + b/a + c/b + d/c + a/d),$$
which by the arithmetic-geometric mean inequality is at least 4.

\textbf{Second solution:}
Choose $A,B,C,D$ as in the first solution.
Note that both the hyperbolas and the area of the convex hull of $ABCD$ are invariant under the transformation $(x,y) \mapsto (xm, y/m)$ for any $m>0$. For $m$ small, the counterclockwise angle from the line $AC$ to the line $BD$ approaches 0; for $m$ large, this angle approaches $\pi$. By continuity, for some $m$ this angle becomes $\pi/2$, that is, $AC$ and $BD$ become perpendicular. The area of $ABCD$ is then $AC \cdot BD$.

It thus suffices to note that $AC \geq 2 \sqrt{2}$ (and similarly for $BD$).
This holds because if we draw the tangent lines to the hyperbola $xy=1$ at the points $(1,1)$ and $(-1,-1)$, then $A$ and $C$ lie outside the region between these lines. If we project the segment $AC$ orthogonally onto the line $x=y=1$, the resulting projection has length at least $2 \sqrt{2}$, so $AC$ must as well.

\textbf{Third solution:}
(by Richard Stanley)
Choose $A,B,C,D$ as in the first solution. Now fixing $A$ and $C$, move $B$ and $D$ to the points at which the tangents to the curve are parallel to the line $AC$. This does not increase the area of the quadrilateral $ABCD$ (even if this quadrilateral is not convex).

Note that $B$ and $D$ are now diametrically opposite; write $B = (-x, 1/x)$ and $D = (x, -1/x)$. If we thus repeat the procedure, fixing $B$ and $D$ and moving $A$ and $C$ to the points where the tangents are parallel to $BD$, then $A$ and $C$ must move to $(x, 1/x)$ and $(-x,-1/x)$, respectively, forming a rectangle of area 4.

\textbf{Remark:}
Many geometric solutions are possible. An example suggested by David Savitt (due to Chris Brewer): note that $AD$ and $BC$ cross the positive and negative $x$-axes, respectively, so the convex hull of $ABCD$ contains $O$. Then check that the area of triangle $OAB$ is at least 1, et cetera.