Calculus

Note that $2^{|x|} - \left|2^{x-1}-1\right| = \begin{cases} 2^{-x} - 1 + 2^{x-1} & \text{for }x \in (-\infty, 0) \\ 2^x - 1 + 2^{x-1} & \text{for } x \in [0, 1) \\ 2^x - 2^{x-1} + 1 & \text{for } x \in [1, \infty) \end{cases}$

For $x \in (-\infty, 0)$ :

$\begin{aligned} 2^{-x} - 1 + 2^{x-1} & = 2^{x-1}+1 \\ 2^{-x} & = 2 \\ \implies x & = \color{#3D99F6} - 1 = a \end{aligned}$

For $x \in [0,1)$ :

$\begin{aligned} 2^{x} - 1 + 2^{x-1} & = 2^{x-1}+1 \\ 2^x & = 2 & \small \color{#D61F06} \text{Since }x < 1 \text{, there is no solution.} \end{aligned}$

For $x \in [1, \infty)$ :

$\begin{aligned} 2^x - 2^{x-1} + 1 & = 2^{x-1}+1 \\ 2^x & = 2^x \\ \implies x & \ge \color{#3D99F6} 1 = b \end{aligned}$

$\implies a+b = -1+1 = \boxed{0}$