In Wikipedia's article about the subject, there is a closed expression for the inverse of the cumulative distribution function. I don't really know how to get to this, because I don't know how to find the inverse of something containing the $sgn(x)$ function or the absolute value.
How is it deduced?
$\DeclareMathOperator{sgn}{sgn}$ The function to be inverted is $$ \begin{align} F(x) &= \int_{-\infty}^x \!\!f(u)\,\mathrm{d}u = \begin{cases} \frac12 \exp \left( \frac{x-\mu}{b} \right) & \mbox{if }x < \mu \\ 1-\frac12 \exp \left( -\frac{x-\mu}{b} \right) & \mbox{if }x \geq \mu \end{cases} \\ &= \tfrac{1}{2} + \tfrac{1}{2} \sgn(x-\mu) \left( 1-\exp \left( -\frac{|x-\mu|}{b} \right) \right) \\ &= y \end{align} $$ with $b > 0$. We start using the case distinctions. For $x \ge \mu$: $$ y = 1-\frac12 \exp \left( -\frac{x-\mu}{b} \right) \iff \\ \exp \left( -\frac{x-\mu}{b} \right) = 2(1-y) \iff \\ -\frac{x-\mu}{b} = \ln(2(1-y)) \iff \\ x = \mu - b \ln(2-2y) $$ For $x < \mu$: $$ y = \frac12 \exp \left( \frac{x-\mu}{b} \right) \iff \\ \exp \left( \frac{x-\mu}{b} \right) = 2 y \iff \\ \frac{x-\mu}{b} = \ln(2y) \iff \\ x = \mu + b \ln(2y) $$ We can combine into $$ x = \mu - \sgn(x - \mu) b \ln(1 + \sgn(x - \mu) - \sgn(x - \mu) 2 y) $$ Comparing with Wikipedia: $$ F^{-1}(p) = \mu - b\,\sgn(p-0.5)\,\ln(1 - 2|p-0.5|) $$ The terms are equal, if $\sgn(x - \mu) = \sgn(y-0.5)$. For $x \ge \mu$ we had $$ 0 \le -b \ln(2-2y) \iff \\ 0 \ge \ln(2 - 2y) \iff \\ 2 - 2 y \le 1 \iff \\ 1 - y \le 1/2 \iff \\ -y \le -1/2 \iff \\ y \ge 1/2 $$ For $x < \mu$ we had $$ 0 > x - \mu = b \ln(2y) \iff \\ 0 > \ln(2y) \iff \\ 2y < 1 \iff \\ y < 1/2 $$ So indeed $\sgn(x-\mu) = \sgn(y - 1/2)$.