For anyone who happens to have Pazy's book on hand, on page 102, we have $$\frac{d}{dt}R(\lambda,A)u(t)=\lambda R(\lambda,A)u(t)-u(t)$$ which implies $$R(\lambda,A)u(t)=-\int_{0}^{t}e^{\lambda(t-\tau)}u(\tau)d\tau$$ I suppose I am probably missing something obvious, but why is this implication valid?
For some background, $A$ is a densely-defined operator with $[0,\infty)\subset\rho(A)$, and $$\limsup_{\lambda\to\infty}\lambda^{-1}\log||R(\lambda,A)||\leq 0$$ We are trying to show that for any initial data in the Banach space, $X$, on which $A$ is defined, there can be at most one solution to the Abstract Cauchy Problem for $A$. Accordingly, $u$ here is assumed to solve this problem for initial data $\vec{0}$. Pazy's proof also depends on Lemma 1.1 of this section and a translation argument as well, both of which make sense to me. I am just stuck on the one step above.
This is merely the variation of constant formula :
For any simple Cauchy problem : $v'(t)=Bv(t)+f(t); \; v(0)=x\in X$, where $B$ generates a $C_0$-semigroup $T(t)$ on a Banach space $X$, the solution (when it exists) is given by the classical formula : $$v(t)=T(t)x+\int_0^t T(t-s) f(s)ds.$$ Now, in your case, we have $v(t)=R(\lambda,A)u(t), \; x=v(0)=0$ (since $u(0)=0$), $B=\lambda I$, $T(t)=e^{t \lambda I}=e^{\lambda t}I$ and $ f(t)=-u(t)$.
We can also prove this result as in the proof of variation constant formula. In fact, take $v(\tau)=R(\lambda,A)u(\tau)$ then your fisrt formula is equivalent to $$v'(\tau)-\lambda v(\tau)=-u(\tau).$$ Multiplying by $e^{-\lambda \tau}$, we obtain $$e^{-\lambda \tau}v'(\tau)-\lambda e^{-\lambda \tau}v(\tau)=-e^{-\lambda \tau}u(\tau).$$ Then $$\frac{d}{dt} [e^{-\lambda \tau}v(\tau)]=-e^{-\lambda \tau}u(\tau).$$ Integrating from $0$ to $t$ and using $v(0)=0$, we have $$e^{-\lambda t}v(t)=-\int_0^t e^{-\lambda \tau}u(\tau) d\tau.$$ Thus, $$v(t)=R(\lambda,A)u(t)=-\int_{0}^{t}e^{\lambda(t-\tau)}u(\tau)d\tau.$$