Check the following:
If a point $\lambda\in \sigma(A)$ is not an eigenvalue of $A$ then $\lambda$ is not isolated in $\sigma(A)$.
$\lambda$ is "isolated from" $\sigma(A)^2 \iff \pm\sqrt{\lambda}$ are both "isolated from" $\sigma(A)$. This tells you: $\lambda$ is not isolated from $\sigma(A)^2 \iff $ at least one of $\pm\sqrt\lambda$ is not isolated from $\sigma(A)$. Here "isolated from" just means isolated point except that the point is not necessarily in the set.
The eigenspace $E_\lambda(A^2) = E_{+\sqrt\lambda}(A) + E_{-\sqrt\lambda}(A)$, in particular if the lefthand side is an infinite dimensional space one of the spaces on the righthand side is infinite dimensional.
Now make use of $\sigma(A^2)=\sigma(A)^2$.
A point $\lambda$ is in $\sigma_{ess}(A^2)$ if and only if it is either not an eigenvalue of $A^2$ (in which case both of $\pm\sqrt\lambda$ are not an eigenvalue of $A$), it is not isolated from $\sigma(A^2)$ (in which case one of $\pm\sqrt\lambda$ is not isolated from $\sigma(A)$), or its eigenspace is infinite dimensional (in which case one of the eigenspaces of $\pm\sqrt\lambda$ must be infinite dimensional). This gives $\sigma_{ess}(A^2)\subseteq \sigma_{ess}(A)^2$.
A point $\lambda$ is in $\sigma_{ess}(A)^2$ if and only if either one of the roots $\pm\sqrt\lambda$ is not an eigenvalue (in which case that root is not isolated in $\sigma(A)$, hence its square $\lambda$ is not an isolated point of $\sigma(A^2)$), one of the roots is not isolated in $\sigma(A)$ (in which case the square also is not isolated), or the eigenspace to $A$ of one of the roots is infinite dimensional (in which case the eiegenspace to $A^2$ of $\lambda$ is also infinite dimensional). This gives $\sigma_{ess}(A)^2\subseteq \sigma_{ess}(A^2)$.
Your argument is fine. The numbers you use are what is known as the numerical range of an operator. It is a basic fact that the spectrum of an operator is contained in the closure of the numerical range. So you show that the numerical range is contained in $[a,1+b]$, which then implies that $\sigma(L_1+L_2)\subset[a,1+b]$.
The inclusion $\sigma(A+B)\subset\sigma(A)+\sigma(B)$ for commuting $A,B$ holds, but equality usually fails. I can only see the abstract of the second paper you mention; what I can say is that the equality in the abstract is not true in general (note in any case that the paper uses the union of the spectra and not their sum). Consider
$$
A_0=\begin{bmatrix} 1&0\\0&0\end{bmatrix},\qquad\qquad B_0=\begin{bmatrix} 3&0\\0&0\end{bmatrix}.
$$
Then $\sigma(A_0)=\{0,1\}$, $\sigma(B_0)=\{0,3\}$, and $\sigma(A_0+B_0)=\{0,4\}$. This is neither $\sigma(A_0)+\sigma(B_0)=\{0,1,3,4\}$ nor $\sigma(A_0)\cup\sigma(B_0)=\{0,1,3\}$. You do have $\sigma(A_0+B_0)\subset\sigma(A_0)+\sigma(B_0)$.
You could also consider
$$
A_1=\begin{bmatrix} 1&0\\0&0\end{bmatrix},\qquad\qquad B_1=\begin{bmatrix} 0&0\\0&3\end{bmatrix}.
$$
Then $\sigma(A_1)=\{0,1\}$, $\sigma(B_1)=\{0,3\}$, and $\sigma(A_1+B_1)=\{1,3\}$. This is not $\sigma(A_0)+\sigma(B_0)=\{0,1,3,4\}$; but you do have $$\sigma(A_1+B_1)\setminus\{0\}=\{1,3\}=(\sigma(A_1)\cup\sigma(B_1))\setminus\{0\}.$$
These examples can be easily turned into examples about the essential spectrum by considering $A=\bigoplus_{n\in\mathbb N}A_0$ and $B=\bigoplus_{n\in\mathbb N}B_0$ and similarly for $A_1$ and $B_1$.
What the examples show is the fact that the spectrum will never tell you the "position" of the operator. A selfadjoint operator is described by its spectrum together with its spectral projections (or spectral measure in infinite-dimension). Two selfadjoint operators that commute will have the same spectral projections, but the location of the eigenvalues can be different, as the two examples show. This leads to different possibilities for the spectrum of the sum.
Best Answer
I find it easiest to prove the contrapositive, $\lambda \in \rho(A)$ if and only if no such sequence $(\psi_n)$ exists.
First we observe that no such sequence $(\psi_n)$ exists if and only if there is a $\delta > 0$ such that $$\lVert (A - \lambda)x\rVert \geqslant \delta \lVert x\rVert$$ for all $x \in \mathcal{H}$.
Next, this form of the condition immediately implies that $A - \lambda$ is injective, and it quickly implies that the range of $A - \lambda$ is closed. Conversely, the open mapping theorem implies that if $A - \lambda$ is injective with closed range, then such a $\delta$ exists.
It remains to be seen that for a self-adjoint $A$, if $A - \lambda$ is injective with closed range, then it is surjective. That follows since $A - \lambda$ is normal (and thus this criterion holds more generally for normal operators, since $A - \lambda$ is normal iff $A$ is normal). Thus suppose $x \in \bigl(\operatorname{im} (A - \lambda)\bigr)^{\perp}$. We then have $$\lVert (A - \lambda)x\rVert^2 = \langle (A - \lambda)x, (A - \lambda)x\rangle = \langle (A - \lambda)^{\ast}(A - \lambda) x, x\rangle = \langle (A - \lambda)(A - \lambda)^{\ast} x, x\rangle = 0$$ by definition of the adjoint and normality. Since by assumption $A - \lambda$ is injective it follows that $x = 0$, hence $\operatorname{im} (A - \lambda)$ is dense, and since it is by assumption closed, that $A - \lambda$ is surjective.