Minimize $|a-1|^3+|b-1|^3$ with constant product $ab=s$

Since you have established that the signs of $a - 1$ and $b - 1$ are identical, one approach would be to find the extrema of $(a - 1)^3 + (b-1)^3$ and then remove those where $a - 1$ and $b - 1$ are of opposite signs.

Plugging in $b = {s \over a}$ directly, you are finding an extremum of the expression $$f(a) = (a - 1)^3 + ({s \over a} - 1)^3$$ So the goal becomes to find a $a$ for which $f'(a) = 0$. If you do the algebra, $f'(a) = 0$ at some $a$ satisfying $$(a - \sqrt{s})(a^2 + (\sqrt{s} - 1)a + s) = 0$$ So you have two possibilities, either $a = b = \sqrt{s}$, or $a$ and $b$ are the two roots of the quadratic equation $x^2 + (\sqrt{s} - 1)x + s = 0$. You can then plug these two possibilities into the expression $(a - 1)^3 + (b - 1)^3 $ to compare the two situations.

While this may look unpleasant to do, since $(a - 1)^3 + (b - 1)^3$ is a symmetric polynomial $(a^3 + b^3) - 3(a^2 + b^2) + 3(a + b) - 1$, you will end out getting a polynomial in $\sqrt{s}$ using $a + b = 1 - \sqrt{s}$ and $ab = s$. You seem to have already computed it to be $1 - 3s - 2s^{3 \over 2}$.