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You should look at Carl Pomerance's followup paper: Ruth-Aaron pairs revisited, . In his first paper with Erdos they proved a result which showed that the number of RA pairs had asymptotic density 0, but just barely. In the followup Pomerance shows that the the sum of the reciprocals converges (which is muchj stronger).

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Apologies that this isn't a complete answer. The condition $\sigma(n+1)=\sigma(n)+1$ means that $\sigma(n),\sigma(n+1)$ are relatively prime. In the question, you've taken care of the divisibility by $2$ part.

Let $p$ be an odd prime and $q>p$ prime. Let $a_{q,p}=a_q=p$ if $q\equiv 1 \bmod p$ and $a_{q,p}=a_q$ equals the order of $q$ modulo $p$ otherwise. If $\nu_q(n)$, the highest power of $q$ dividing $n$ is, $-1$ mod $a_q$, then $p$ divides $\sigma(n)$. So it is necessary that if $q_1,q_2$ are some primes, then one musn't simultaneously have $\nu_{q_1}(n)\equiv -1 \bmod a_{q_1,p}$ and $\nu_{q_2}(n+1)\equiv -1 \bmod a_{q_2,p}$ for any $p,q_1,q_2$.

For example, if $n=q^2$, where $q$ is prime and $1$ mod $3$, then $n+1$ is $2\bmod 4$ and $3$ divides both $\sigma(n), \sigma(n+1)$, so this is an example of $n$ that must be excluded.

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