Can a sequence of finitely many elementary operations/algorithm give a specific closed form for a recurrence relation?

This is not an answer, but is too long for a comment. Below is a natural, ansatz-free method that surprisingly leads to the double of the wished result. I suspect that with a little more ingenuity, it can lead to a full solution.

So, we start with the recurrence relation

$$ a_n=1 + \left(\dfrac{n+1}{2n}\right) a_{n-1} \tag{$A_0$} $$

Step 1 : we are optimistic, and assume that $(a_n)$ has a limit, in other words that there has an expansion of the form $a_n=l+o(1)$. Passing to the limit in $(A_0)$ above, we see that $l=2$. This suggests the rescaling $a_n=2+b_n$, which yields

$$ b_n=\dfrac{1}{n} + \left(\dfrac{n+1}{2n}\right) b_{n-1} \tag{$B_0$} $$

Step 2 : we are optimistic, and assume that $(a_n)$ has a Taylor expansion of the form $a_n=2+\dfrac{l}{n}+o(\dfrac{1}{n})$. Then $l$ is the limit of the sequence $a_n^1=nb_n$, so let us see what recurrence relation is satisfied by this new sequence. It turns out to be

$$ a^1_n=1 + \left(\dfrac{n+1}{2(n-1)}\right)a^1_{n-1} \tag{$A_1$} $$

Step $3$: Passing to the limit in $(A_1)$ above, we see that the limit of $a^1_n$ is $2$ if it exists. This suggests the rescaling $a^1_n=2+b^1_n$, which yields

$$ b^1_n=\dfrac{2}{n-1} + \left(\dfrac{n+1}{2(n-1)}\right) b^1_{n-1} \tag{$B_1$} $$

Step 4: arguing as in step 2, let us now put $a_n^2=(n-1)b^1_n$ (we use $n-1$ rather than $n$, because of the $n-1$ in the denominator of $(B_1)$ above). We deduce

$$ a^2_n=2 + \left(\dfrac{n+1}{2(n-2)}\right)a^1_{n-1} \tag{$A_2$} $$

Iterating this argument, we obtain for every $k\geq 0$,

$$ \begin{array}{lclc} a^k_n &=& k! + \left(\dfrac{n+1}{2(n-k)}\right)a^k_{n-1} & (A_k) \\ b^k_n &=& \dfrac{(k+1)!}{2(n-k)} + \left(\dfrac{n+1}{2(n-k)}\right)b^k_{n-1} & (B_k) \end{array} $$

and we have the successive identities

$$ \begin{array}{lcl} a_n &=& 2+ \dfrac{1}{n}a^1_n \\ &=& 2+ \dfrac{2}{n} +\dfrac{1}{n(n-1)}a^2_n \\ &=& 2+ \dfrac{2}{n} +\dfrac{4}{n(n-1)} +\dfrac{1}{n(n-1)(n-2)}a^3_n \\ &=& \ldots \\ \end{array} $$

At the $r$-th step ($0\leq r \lt n $) of this iteration, we obtain

$$ a_n = \sum_{k=0}^{r}\dfrac{2k!}{n(n-1)\ldots(n-(k-1))} + \dfrac{1}{n(n-1)(n-2)\ldots(n-r)}a^r_n $$

In other words,

$$ a_n = \sum_{k=0}^{r}\dfrac{2}{\binom{n}{k}} + \dfrac{(r+1)!}{\binom{n}{r+1}}a^r_n \tag{1} $$

In particular, when $r=n-1$,

$$ a_n = \sum_{k=0}^{n-1}\dfrac{2}{\binom{n}{k}} + n!a^{n-1} _n \tag{2} $$