jmc

Let $z=x+yi,$ where $x$ and $y$ are real numbers. The given inequality is equivalent to $$|z^2+1|\le 2|z|.$$Then $$|(x^2-y^2+1)+2xyi|\le 2|x+yi|.$$This is equivalent to $|(x^2-y^2+1)+2xyi{|}^2\le 4|x+yi{|}^2,$ so $$(x^2-y^2+1{)}^2+4x^2{y}^2\le 4x^2+4y^2.$$This simplifies to $$x^4+2x^2{y}^2+y^4-2x^2-6y^2+1\le 0.$$We can write this as $$(x^2+y^2{)}^2-2(x^2+y^2)+1-4y^2\le 0,$$or $(x^2+y^2-1{)}^2-4y^2\le 0.$ By difference of squares, $$(x^2+y^2-1+2y)(x^2+y^2-1-2y)\le 0.$$Completing the square for each factor, we get $$(x^2+(y+1{)}^2-2)(x^2+(y-1{)}^2-2)\le 0.$$The factor $x^2+(y+1{)}^2-2$ is positive, zero, or negative depending on whether $z$ lies inside outside, on, or inside the circle $$|z+i|=\sqrt{2}.$$Similarly, the factor $x^2+(y-1{)}^2-2$ is positive, zero, or negative depending on whether $z$ lies inside outside, on, or inside the circle $$|z-i|=\sqrt{2}.$$This tells us that $z$ lies in $S$ if and only if $z$ lies in exactly one of these two circles.



We can divide $S$ into six quarter-circles with radius $\sqrt{2},$ and two regions that are squares with side length $\sqrt{2}$ missing a quarter-circle.



Hence, the area of $S$ is $4\cdot \frac{1}{4}\cdot (\sqrt{2}{)}^2\cdot \pi +2\cdot (\sqrt{2}{)}^2=\boxed{2\pi +4}.$