We can use the integral kernel of $R=(A-z)^{-1}$ and nothing much changes when $p\not= 2$. By the variation of constants formula, $(Rf)(x)=\int_0^1 G(x,t;z) f(t)\, dt$. Here $G$ could of course be worked out explicitly, but I'll just use that $$ G(x,t;z) =\sum_{n=1}^{\infty} \frac{u_n(x)u_n(t)}{\lambda_n-z} , $$ with $u_n(x)=\sqrt{2} \cos n\pi x$ (the $n$th eigenfunction) and $\lambda_n=-n^2\pi^2$ is the corresponding eigenvalue. It doesn't matter here what $p$ is because we always need to solve the same ODE with the same boundary conditions (and, if we want to, we can assume that $f\in C_0^{\infty}(0,1)$).

Now clearly $$ |(Rf)(x)|\le \sum \frac{|u_n(x)|}{|\lambda_n-z|} \|u_n\|_{p'}\|f\|_p , $$ and $\|u_n\|_p, \|u_n\|_{p'}$ are uniformly bounded, so to obtain the desired bound, I only need to show that $$ \sum_{n=1}^{\infty} \frac{1}{|\lambda_n-z|} \lesssim \frac{1}{\textrm{dist}(z,\{\lambda_n\})} . $$ This is clear since $\lambda_n=-n^2\pi^2$.

We can use the integral kernel of $R=(A-z)^{-1}$ and nothing much changes when $p\not= 2$. By the variation of constants formula, $(Rf)(x)=\int_0^1 G(x,t;z) f(t)\, dt$. Here $G$ could of course be worked out explicitly, but I'll just use that $$ G(x,t;z) =\sum_{n=1}^{\infty} \frac{u_n(x)u_n(t)}{\lambda_n-z} , $$ with $u_n(x)=\sqrt{2} \cos n\pi x$ (the $n$th eigenfunction) and $\lambda_n=-n^2\pi^2$ is the corresponding eigenvalue. It doesn't matter here what $p$ is because we always need to solve the same ODE with the same boundary conditions (and, if we want to, we can assume that $f\in C_0^{\infty}(0,1)$).

Now clearly $$ |(Rf)(x)|\le \sum \frac{|u_n(x)|}{|\lambda_n-z|} \|u_n\|_{p'}\|f\|_p , $$ and $\|u_n\|_p, \|u_n\|_{p'}$ are uniformly bounded, so to obtain the desired bound, I only need to show that $$ \sum_{n=1}^{\infty} \frac{1}{|\lambda_n-z|} \lesssim \frac{1}{\textrm{dist}(z,\{\lambda_n\})} . $$ This is clear for $z$ between eigenvalues (or $z<1$, say) since $\lambda_n=-n^2\pi^2$.