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 $n$:
$$a_{n+1}=1+\frac{na_n}{n+a_n}=1+\frac{np_n}{p_n+nq_n}=1+\frac{p_{n+1}-q_{n+1}}{q_{n+1}}=\frac{p_{n+1}}{q_{n+1}}.$$
This answers the second part of the OP's question. More precisely, it would also be desirable to prove that $\gcd(p_n,q_n)=1$, but I have not verified this.

**Added 1.** I can now prove that $\gcd(p_n,q_n)=1$, i.e., the fraction $(3)$ is in lowest terms. Let us proceed by induction. For $n=1$, the statement is clear. We assume therefore that $\gcd(p_n,q_n)=1$ for a given $n$, and we prove that $\gcd(p_{n+1},q_{n+1})=1$. If this is not the case, then there is a prime $\ell$ that divides both $p_{n+1}$ and $q_{n+1}$. From $(1)$ we get $\ell\mid np_n$, hence from $(2)$ we get $\ell\mid n^2 q_n$. As $p_n$ and $q_n$ are relatively prime by the induction hypothesis, this forces $\ell\mid n$, and then from $(2)$ we get $\ell\mid p_n$. However,
$$p_n=1+\sum_{k=1}^{n-1}\binom{n}{k}\binom{n-1}{k}k!=1+n\sum_{k=1}^{n-1}\binom{n-1}{k-1}\binom{n-1}{k}(k-1)!\ ,$$
whence $p_n\equiv 1\pmod{n}$, and therefore $p_n\equiv 1\pmod{\ell}$. This contradiction shows that $\gcd(p_{n+1},q_{n+1})=1$, and we are done.

**Added 2.** I can now answer the remaining first part of the question. Briefly, as observed in the comments, the observed "property of $a$" is equivalent to the identity
$$e E_1(1)=[1/1,1,1/2,1,1/3,1,1/4,1,1/5,1,\dots],$$
where (cf. [Wikipedia][3])
$$E_1(x):=\int_x^\infty\frac{e^{-t}}{t}\,dt,\qquad x>0.$$
More generally, it is known that (cf. [Wikipedia][3])
$$e^x E_1(x)=[x/1,1,x/2,1,x/3,1,x/4,1,x/5,1,\dots].$$

  [1]: https://oeis.org/A000262
  [2]: https://oeis.org/A002720
  [3]: https://en.wikipedia.org/wiki/Exponential_integral