We have,

$f\left(x\right)=\left\{\begin{array}{ll}\sin x-e^x& \mathrm{if} x\le 0\\ a+\left[-x\right]& \mathrm{if} 0<x<1\\ 2x-b& \mathrm{if} x\ge 1\end{array}\right.$

If $f\left(x\right)$ is continuous at $x=a$ then $\underset{x\to a^{-}}{\lim }f\left(x\right)=f\left(a\right)=\underset{x\to a^{+}}{\lim }f\left(x\right)$.

Continuous at $x=0$

$f\left(0^{+}\right)=f\left(0^{-}\right)$

$\Rightarrow a-1=0-e^0$

$\Rightarrow a=0$

Continuous at $x=1$

$f\left(1^{+}\right)=f\left(1^{-}\right)$

$\Rightarrow 2\left(1\right)-b=a+\left(-1\right)$

$\Rightarrow b=2-a+1$

$\Rightarrow b=3$

$\therefore a+b=3$