[The sequence $c_n$ is number A000699 in the OEIS. Also, see Flajolet and Noy, Analytic Combinatorics of Chord Diagrams (INRIA research report #3914), $\S1$, p. 2, and the earlier question From generating functions to recurrence on this sequence.]
Let $I_n:=\{1,\ldots,2n\}$. Suppose that, in a connected arrangement of arcs on $I_n$, we mark one of the segments $[i,i+1]$, where $i\in\{1,\ldots,2n-1\}$. Call the number of marked arrangements $d_n$; obviously $d_n=(2n-1)c_n$. Let $B:=\sum_{n>0} d_n x^n$ be the generating function for $d_n$.
By the argument you give above, given a connected arrangement of arcs on $I_n$, after removing $\{1,a\}$, we get a sequence of connected arrangements of arcs on subsets of $I_n$, and if we renumber the vertices of each arrangement in an order-preserving manner, we get a sequence $X_1$, $\ldots$, $X_k$ of connected arrangements of arcs on $I_{n_1}$, $\ldots$, $I_{n_k}$, where $n_1+\cdots+n_k=n-1$. Since the sequence we started with is nested, for $i=1$, $\ldots$, $k-1$, we can mark the segment in each $I_{n_i}$ which, in the original arrangement, would be expanded to contain $X_{i+1}$; in the case $i=k$, mark the segment which is expanded to contain $a$. This gives a sequence $X'_1$, $\ldots$, $X'_k$ of marked connected arrangements of arcs on $I_{n_1}$, $\ldots$, $I_{n_k}$ from which the original connected arrangement of arcs on $I_n$ can be recovered. Therefore, there is a bijective correspondence between
- connected arrangements of arcs on $I_n$
and
- sequences of marked connected arrangements of arcs on $I_{n_1}$, $\ldots$, $I_{n_k}$, where $n_1+\cdots+n_k=n-1$. ($k$ may be zero, in which case the sequence is empty.)
So, $ c_n=\sum_{k\ge 0} \sum_{n_1+\cdots+n_k=n-1} d_{n_1}\cdots d_{n_k}.\qquad (1) $ To turn (1) into an equality of generating functions, multiply each side of (1) by $x^n$ and sum over $n$. Doing this to the left-hand side gives $A$, the generating function for $c_n$. If we fix some $k$ in the right-hand side and then perform the multiplication and summation, we get $ \sum_{n>0} \sum_{n_1+\cdots+n_k=n-1} d_{n_1}\cdots d_{n_k} x^{n} $ $ = \sum_{n>0} \sum_{n_1+\cdots+n_k=n-1} d_{n_1}\cdots d_{n_k} x^{n_1+\cdots+n_k+1} $ $ = x \sum_{n_1,\ldots,n_k} d_{n_1}\cdots d_{n_k} x^{n_1+\cdots+n_k}, $ which is equal to $ x B^k. $ Therefore, reinserting the summation over $k$ gives $ A=x \sum_{k\ge 0} B^k.\qquad (2) $ But, since $ A=\sum_{n>0} c_n x^n $ and $2xA^\prime = \sum_{n>0} 2nc_n x^n$ we have $2xA^\prime-A = \sum_{n>0} (2n-1)c_nx^n = \sum_{n>0} d_n x^n=B.\qquad (3)$ Combining (2) and (3) gives $ A=x \sum_{k\ge 0} (2xA'-A)^k, $ as desired.