I have that:
$$S_n=\frac{b\cdot S_{n-1}}{b+S_{n-1}}+a$$
where $n\in\mathbb{N}^+$ and $S_1=a+b$.
So, for $S_2$ we will get:
$$S_2=\frac{b\cdot S_{2-1}}{b+S_{2-1}}+a=\frac{b\cdot S_1}{b+S_1}+a=\frac{b\cdot(a+b)}{b+(a+b)}+a$$
And for $S_3$:
$$S_3=\frac{b\cdot S_{3-1}}{b+S_{3-1}}+a=\frac{b\cdot S_2}{b+S_2}+a=\frac{b\cdot\left(\frac{b\cdot(a+b)}{b+(a+b)}+a\right)}{b+\left(\frac{b\cdot(a+b)}{b+(a+b)}+a\right)}+a$$
And so on. Now my question is: what happens when $n\to\infty$?
Assume that $S_n = \frac{p_n}{q_n}$ is associated with $(p_n,q_n)\in\mathbb{R}^2$. The recurrence
$$ S_{n} = \frac{(a+b)S_{n-1}+ab}{S_{n-1}+b}\tag{1} $$
can be written in the following form
$$ \begin{pmatrix}p_n \\ q_n \end{pmatrix} = \begin{pmatrix} a+b & ab \\ 1 & b\end{pmatrix} \begin{pmatrix}p_{n-1} \\ q_{n-1} \end{pmatrix} \tag{2}$$
hence the closed form of both the $\{p_n\}_{n\geq 0}$ and the $\{q_n\}_{n\geq 0}$ sequences just depends on the diagonalization of $M=\begin{pmatrix} a+b & ab \\ 1 & b\end{pmatrix}$, whose eigenvalues are given by $\zeta_{\pm}=\frac{(a+2b)\pm\sqrt{a^2+4ab}}{2}$.
In particular $$ S_n = \frac{ A \zeta_+^n + B \zeta_-^n}{C \zeta_+^n + D \zeta_-^n}\tag{3}$$ for some constants $(A,B,C,D)$ depending on the initial values. By taking the limit as $n\to +\infty$, we simply get $\color{red}{\frac{A}{C}}$. On the other hand, by assuming $S_n\to L$ as $n\to \infty$, such limit has to fulfill $$ L = a + \frac{bL}{b+L} \tag{4}$$ hence the only chances are $L=\frac{1}{2}\left(a\pm\sqrt{a^2+4ab}\right)$.