Laplace transform of a finite duration signal
The property claimed by Schaum and Oppenheim is also true for the given example. Note that the Laplace transform

$$X(s)=\frac{1-e^{-5(s+2)}}{s+2}\tag{1}$$

has no pole at $s=-2$:

$$\lim_{s\to -2}X(s)=\frac{1-(1-5(s+2))}{s+2}\Big{|}_{s=-2}=5$$

So the ROC is indeed the entire $s$-plane. Even though the ROCs of the two individual signals in your solution have the same right half-plane as ROC, their sum has the entire $s$-plane as ROC because the two signals cancel everywhere except in a finite interval.