Let $A$ be a bounded non-normal operator on a separable Hilbert space and $z\in\mathbb{C}$ such that $\delta:=\mathrm{dist}(z,\sigma(A))>0$.
Is it true, like the case of self-adjoint operators, that:
1) $$ \|(A-zI)^{-1}\|\leq \frac{1}{\delta} $$
2) If $B$ is a bounded operator such that $\|B\|<\delta$, then $$ \|(A+B-zI)^{-1}\| < \frac{1}{\delta-\|B\|}$$
If it is false, please provide a counter-example.
Look at the case of a $2 \times 2$ nilpotent matrix to disprove your conjecture. Such a matrix has spectrum consisting of one point, namely 0. More explicitly, look at $$ A := \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}. $$ Then $A^2=0$, which gives a truncated geometric series for the resolvent: $$ (\lambda I-A)^{-1}= \frac{1}{\lambda}I+\frac{1}{\lambda^2}A. $$ You cannot bound the norm of the above by a constant times $1/|\lambda|$ near $\lambda=0$ because of the $1/\lambda^2$ term.