Here's a really elementary proof (which is a slight modification of Fanfan's answer to a question of mine). As Calle shows, it is easy to see that the eigenvalue $1$ is obtained. Now, suppose $Ax = \lambda x$ for some $\lambda > 1$. Since the rows of $A$ are nonnegative and sum to $1$, each element of vector $Ax$ is a convex combination of the components of $x$, which can be no greater than $x_{max}$, the largest component of $x$. On the other hand, at least one element of $\lambda x$ is greater than $x_{max}$, which proves that $\lambda > 1$ is impossible.
A first step could be the following: A corollary to Weyl's eigenvalue inequalities (see, e.g., Horn & Johnson, second edition, Corollary 4.3.15), is that the largest eigenvalue of a sum of matrices is as least as large as the sum of the largest eigenvalue of one of the matrices and the smallest eigenvalue of the other matrix, i.e.,
$$ \lambda_M(R_1+R_2)\ge\lambda_M(R_1)+\lambda_1(R_2) $$
Since the matrices are covariance matrices, they are positive semidefinite, and hence $\lambda_1(R_2)\ge0$.
Note that you can easily omit the identity matrix by just adding it to the matrix $R_2$. Note further that, since all matrices involved are symmetric,
$$ R_1(R_1+R_2)^{-2}R_1 = R_1(R_1+R_2)^{-1}(R_1+R_2)^{-1}R_1 = R_1(R_1+R_2)^{-1}\left(R_1(R_1+R_2)^{-1}\right)^T. $$
Hence, for the moment it might suffice to determine the largest eigenvalue of $R_1(R_1+R_2)^{-1}$ and then use the Corollary to Gelfand's formula to bound
$$ \lambda_M\left(R_1(R_1+R_2)^{-1}\left(R_1(R_1+R_2)^{-1}\right)^T\right) \le \lambda_M(R_1(R_1+R_2)^{-1})^2. $$
Best Answer
So, let
$$P=\mathrm{diag}_i\left(\dfrac{1}{\sqrt{\sigma_i}}\right)$$
where $\sigma_i=\sum_{j=1}^ns_{ij}$.
We need to check the maximum eigenvalue of $PSP$. The eigenvalues are the same as
$$ SP^2=S\mathrm{diag}_i\left(\dfrac{1}{\sigma_i}\right). $$
Now we have that
$$ S\mathrm{diag}_i\left(\dfrac{1}{\sigma_i}\right)\sigma=S\mathrm{1}=\sigma. $$
Since $SP^2$ is a positive matrix, or at least nonnegative, and we have that at all diagonal entries are equal to 1, then $\sigma$ is a positive vector and, by virtue of the Perron-Frobenius theorem, this implies that 1 is the dominant eigenvalue of $S$.