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
A self-adjoint operator $T$ admitting a cyclic vector is, as you said, necessarily the multiplication operator on $L^2(\sigma(T),\mu)$ given by $$ T(f)x=xf(x). $$ So, once $\sigma(T)$ is fixed, all you have to determine is the measure $\mu$ which necessarily has full support.
It is also not hard to see that two equivalent, that is, mutually absolutely continuous measures lead to unitarily equivalent operators.
In the special case of compact operators, the spectrum always contains zero and is discrete, except for the fact that zero may be an accumulation point. In this case two measures $\mu_1$ and $\mu_2$ with full support are equivalent iff they both assign zero for $\{0\}$ (in which case the associated operators are both injective) or they both do not (in which case both operators have a one-dimensional kernel).
As a consequence, if two injective, self-adjoint, compact operators admit cyclic vectors and share spectrum then, yes, they are unitarily equivalent.
On the other hand suppose you take the counting measure $\mu_1$ on the set $$X=\{0\}\cup\{1/n:n\in\mathbb N\}$$ and let $\mu_2$ be the same measure, except that $\mu_2$ assigns zero to $\{0\}$. Then the corresponding operators will provide an example to the first paragraph in your question in which $T_2$ is injective and compact.