Let a Banach space $X$ and $A \colon D(A) \to X$ a linear operator. Let $R_\lambda$ its resolvent operator. Then we have the equality $A R_\lambda x= R_\lambda A x$ for all $x \in D(A)$.
Now assume that $A$ is the generator of a $\mathcal C_0-$semigroup $S(t)$. Analogously, is it true that $$ R_\lambda S(t)=S(t)R_\lambda ?$$
Yes. Indeed, define, for a fixed $x\in D(A)$, the function $f(s)=S(t-s)R_\lambda S(s)x$ in the interval $[0,t]$. Then show that $f$ is differentiable, and $f'(s)=0$ in $[0,t]$ (by the equality you mentioned). Hence, $f(t)=f(0)$ implies that $R_\lambda S(t)=S(t)R_\lambda$ in $D(A)$, and then in $X$ by density.