For $n\geq 1$, let $p_n$ be the $(n+1)$-th term of [A000262][1], and let $q_n$ be $n$-th term of [A002720][2]. Then, according to the description of these two sequences (more precisely by the contributions of Dennis P. Walsh and Paul Berry)
$$p_n=\sum_{k=0}^{n-1}\binom{n}{k}\binom{n-1}{k}k!\qquad\text{and}\qquad
q_n=\sum_{k=0}^{n-1}\binom{n-1}{k}^2k!\ .$$
We derive some identities. First,
\begin{align}p_{n+1}-q_{n+1}&=\sum_{k=0}^{n}\binom{n+1}{k}\binom{n}{k}k!-\sum_{k=0}^{n}\binom{n}{k}^2k!\\[6pt]
&=\sum_{k=0}^n\left(\binom{n+1}{k}-\binom{n}{k}\right)\binom{n}{k}k!\\[6pt]
&=\sum_{k=1}^n\binom{n}{k-1}\binom{n}{k}k!\\[6pt]
&=n\sum_{k=1}^n\binom{n}{k-1}\binom{n-1}{k-1}(k-1)!\\[6pt]
&=n\sum_{k=0}^{n-1}\binom{n}{k}\binom{n-1}{k}k!\\[6pt]
&=np_n.
\end{align}
That is,
$$p_{n+1}=np_n+q_{n+1}.\tag{1}$$
Second,
\begin{align}q_{n+1}-p_n&=\sum_{k=0}^{n}\binom{n}{k}^2k!-\sum_{k=0}^{n-1}\binom{n}{k}\binom{n-1}{k}k!\\[6pt]
&=n!+\sum_{k=0}^{n-1}\binom{n}{k}\left(\binom{n}{k}-\binom{n-1}{k}\right)k!\\[6pt]
&=n!+\sum_{k=1}^{n-1}\binom{n}{k}\binom{n-1}{k-1}k!\\[6pt]
&=n!+n\sum_{k=1}^{n-1}\binom{n-1}{k-1}^2(k-1)!\\[6pt]
&=n!+n\sum_{k=0}^{n-2}\binom{n-1}{k}^2k!\\[6pt]
&=n\sum_{k=0}^{n-1}\binom{n-1}{k}^2k!\\[6pt]
&=nq_n.\end{align}
That is,
$$q_{n+1}=p_n+nq_n.\tag{2}$$
From $(1)$ and $(2)$, it follows by induction that 
$$a_n=\frac{p_n}{q_n}.\tag{3}$$ Indeed, $(3)$ holds for $n=1$. Assuming $(3)$ holds for a given $n$, it also holds for $n+1$ in place of $n$, since
$$a_{n+1}=1+\frac{na_n}{n+a_n}=1+\frac{np_n}{p_n+nq_n}=1+\frac{p_{n+1}-q_{n+1}}{q_{n+1}}=\frac{p_{n+1}}{q_{n+1}}.$$
This answers the second part of the OP's question. More precisely, it would also be desirable to prove that $\gcd(p_n,q_n)=1$. This seems straightforward along similar lines, but I have not verified it (for lack of time).

  [1]: https://oeis.org/A000262
  [2]: https://oeis.org/A002720