Let $A$ be the generator of a $C_0$-semigroup $(T(t))_{t \geq 0}$ of contractions on a Banach space $X$ and $B \in \mathcal L(X)$ a bounded operator. To apply some approximation formula I want to show that there exist $\lambda > 0$ such that $\lambda - (A + B)$ has dense range, i.e., $X= \overline{\operatorname{ran}}(\lambda - (A + B))$ and I am not quite sure how to do that.
To begin with, I know that since $(T(t))_{t \geq 0}$ is a contraction semigroup, $\lambda - A$ is invertible for each $\lambda > 0$. Moreover, it holds $$ R(\lambda, A + B) = R(\lambda, A) (I - BR(\lambda, A + B)) \tag{1}$$ for each $\lambda \in \rho(A + B)$. Since $\Vert R(\lambda, A + B)) \Vert \to 0$ as $\vert \lambda \vert \to \infty$, one should be able to find $\lambda > 0$ such that $\Vert BR(\lambda, A + B) \Vert < 1$. Hence, both operators $\lambda - A$ and $BR(\lambda, A + B)$ should be invertible for some $\lambda > 0$ that is large enough. Hence, due to $(1)$ $\lambda - (A + B)$ should be invertible as long as I can find a sequence $(\lambda_n)_{n \in \mathbb N}$ in $\rho(A + B)$ with $\lambda_n \to \infty$ as $n \to \infty$. In this case, the range of $\lambda - (A + B)$ would be $X$ for some $\lambda > 0$ big enough. But I am not sure how I can deduce the existence of such a sequence from scratch.
On the other hand, I think my argument is to difficult. Because it is a consequence of the well-known Hille-Yosida theorem that $A+ B$ generates a semigroup of type $(1, \Vert B \Vert)$. In this case, one would know that $\lambda - (A + B)$ is invertible for each $\lambda > \Vert B \Vert$, and therefore has dense range, by standard semigroup theory. But I hoped to avoid this consequence of Hille-Yoshida in my proof for sake of simplicity.
So my question is, if there is any chance to show that $\lambda - (A + B)$ has dense range in this setting without Hille-Yosida, maybe by results from spectral theory, for example?
We have the following small perturbations lemma (which I'll prove here in your special case, see e.g. Theorem 11.1.3 in "Linear Operators and their Spectra" by Davis for a more general result).
Proof of Lemma: Let $R = R(\lambda, A)(\operatorname{Id} - cBR(\lambda,A))^{-1}$ (which exists by a von Neumann series argument). Then \begin{align*} (\lambda - (A+cB))R &= ((\lambda - A) - cB) R(\lambda, A)(\operatorname{Id} - cBR(\lambda,A))^{-1} \\ &= (\operatorname{Id} - cBR(\lambda,A))((\operatorname{Id} - cBR(\lambda,A))^{-1} \\&= \operatorname{Id} \end{align*} $\square$
To conclude, we'd like to be able to show that we can find $\lambda \in \rho(A)$ such that we can plug $c = 1$ into the above. Since $B$ is bounded, this is straightforward. Indeed, $$\|BR(\lambda,A)\| \leq \|B\| \|R(\lambda,A)\| \to 0$$ as $\lambda \to \infty$ with $\lambda \in \mathbb{R}$ (where I use that $(0, \infty) \subseteq \rho(A)$). So for $\lambda \in \mathbb{R}$ large enough, $\|BR(\lambda,A)\|^{-1}>1$ and so we can take $c = 1$ in the lemma to see that $\lambda \in \rho(A+B)$ and hence $\lambda - (A+B)$ is surjective.