Answer: 66

Solution:

Let $f(t) = \bigl|\,||t| - 2| - 1\,\bigr|,$ so the equation is $f(x) + f(y) = 1.$ The function $f$ is even, and for $t \ge 0:$ on $[0, 2],$ $||t| - 2| - 1 = (2 - t) - 1$ $= 1 - t,$ so $f(t) = |t - 1|;$ on $[2, 4],$ $f(t) = |t - 3|;$ and for $t \gt 4,$ $f(t) = t - 3 \gt 1,$ which is too large. So on the relevant range, $f(t) = |t - a|$ where $a \in \{-3, -1, 1, 3\}$ is the nearest of those four values to $t.$

Therefore $S$ is the union of the $16$ taxicab circles $$\begin{aligned} &|x - a| + |y - b| = 1, \\ &\qquad a, b \in \{-3, -1, 1, 3\}, \end{aligned}$$ which meet only at isolated points. Each is a square (diamond) with diagonal $2,$ hence side $\sqrt{2}$ and perimeter $4\sqrt{2}.$

The total length is $16 \cdot 4\sqrt{2} = 64\sqrt{2},$ so $a + b = 64 + 2 = 66.$