For ease of notation I will denote $\mathcal N(\lambda-A)$ by $N$ and $\mathcal N(\lambda-A)^\perp$ by $N'$. You then have $H=N\oplus N'$ and the operator $A$ respects this decomposition as $A$ is self-adjoint, so you can write $A=(\lambda\Bbb 1\lvert_{N})\oplus B$ where $B:N'\to N'$. The operator $B$ is completely arbitrary, so long as it is self-adjoint and doesn't have an eigenvalue at $\lambda$.
You now want conditions on $A$ so that the operator $B-\lambda$ is invertible. The conditions you list, like for example $\lambda -A$ being non-negative, are not sufficient to guarantee this. As an example consider $\lambda=0$ and $H=L^2([0,1],dx)\oplus \Bbb R$ and let $A(f,z) = (x\cdot f,0)$. This is a non-negative operator and the operator $B$ is the multiplication with $x$ operator on $L^2([0,1],dx)$, which is not invertible.
You can continue to play this game, finding for a condition $A$ a suitable, completely arbitrary operator $B$ (except for the conditions of being self-adjoint and not having an eigenvalue $\lambda$).
So when is $B$ invertible? The relevant criterium is the following: $B$ is invertible iff $\lambda$ is an isolated point of $\sigma(A)$.
This follows directly from
$$\sigma(B)=\overline{\sigma(A)-\{\lambda\}}.$$
So we should think about why that equation is true. First note that in our decomposition $A=(\lambda\Bbb1\lvert_N)\oplus B$ you have $\tilde\lambda-A= (\tilde \lambda- \lambda)\Bbb1\lvert_{N}\oplus (\tilde \lambda -B)$. This is invertible if and only if both operator summands are invertible, so for $\tilde\lambda\neq\lambda$ you have that $\tilde\lambda -A$ is invertible iff $\tilde\lambda -B$ is invertible, in particular $\sigma(B)$ can differ from $\sigma(A)$ by at most the point $\lambda$. However $\sigma(B)$ must be closed, so if $\lambda$ is not an isolated point of $\sigma(A)$ it must be in $\sigma(B)$. On the other hand if $\lambda$ were a isolated point of $\sigma(A)$ and $\lambda\in \sigma(B)$, then $\lambda$ must be isolated in $\sigma(B)$. But isolated points in the spectrum correspond to eigenvalues, so $B$ would have to have an eigenvalue at $\lambda$, which is forbidden, hence if $\lambda$ is isolated it is not in $\sigma(B)$.
Best Answer
This is true. It might be true under more general assumptions on the space $\mathcal N(\lambda-A)^\perp$, but the fact that this is the complement of an eigenspace makes the relevant calculation very simple here.
Let $z\in \mathcal D(A\lvert_{\mathcal N(\lambda-A)^\perp}^*)$, that is $z\in\mathcal N(\lambda-A)^\perp$ so that $|\langle z , Ax\rangle|≤\|x\|\,C_z$ for all $x\in\mathcal D(A)\cap \mathcal N(\lambda -A)^\perp$. We want to show that $z\in\mathcal D(A\lvert_{\mathcal N(\lambda-A)^\perp})$. This, together with symmetry, shows self-adjointness of $A\lvert_{\mathcal N(\lambda-A)^\perp}$.
Now for any $y\in\mathcal D(A)$ you have $y=x+v$ with $v\in \mathcal D(A)\cap \overline{\mathcal N(\lambda-A)}= \mathcal N(\lambda -A)$, that is $v$ an eigenvector of $A$ to the eigenvalue $\lambda$, and $x\in \mathcal D(A)\cap\mathcal N(\lambda-A)^\perp$. Note that $\|x+v\|^2=\|x^2\|+\|v\|^2$ since $x$ and $v$ are perpendicular. It follows that:
$$\langle z, Ay\rangle = \langle z,Ax\rangle +\lambda\langle z,v\rangle = \langle z,Ax\rangle$$ since $z$ is perpendicular to the eigenvectors to $\lambda$ of $A$. Hence $$|\langle z, Ay\rangle| ≤ \|x\|\,C_z≤\|y\|\,C_z$$ and $z\in\mathcal D(A^*)$. But $\mathcal D(A^*)=\mathcal D(A)$, hence $z\in \mathcal D(A\lvert_{\mathcal N(\lambda-A)^\perp})$ and $A\lvert_{\mathcal N(\lambda -A)^\perp}$ is self-adjoint.