The perturbation theorem is stated as follows:
Let $A$ be a stable family of infinitesimal generators with stability constants $M$ and $\omega$. Let $B$ be bounded linear operators on $X$. If $\|B\|<K$, then $A+B$ is a stable family of infinitesimal generators with stability constants $M$ and $\omega + M\,K$.
Is for the operator $A-B$, the second stability constant is still $\omega +M\,K$ or $\omega-M\,K$?
Remark that $\|-B\|=\|B\|<K$. Now, applying the result to $-B$ yields that the constant for $A+(-B)=A-B$ is also $\omega +M\,K$.