Define the function
$$ T_a(x) := 1/(a + 1/x). \tag1 $$
It is easy to check that
$$ T_{a+b}(x) = T_a (T_b(x)). \tag2 $$
Specialize this equality to get
$$ x = T_ 0(x) = T_ 1(T_{-1}(x)) = T_{-1}(T_ 1(x)). \tag3 $$
Equivalently, this implies that
$$ z = 1/(-1 + 1/y) \;\; \text{iff} \;\; y = 1/(1 + 1/z) \;\; \text{iff} \;\; z - y = y\,z. \tag4 $$
Expand, using formal power series to get
$$ z\!=\!y + y^2 + y^3 + y^4 + \cdots \;\; \text{iff} \;\; y\!=\!z - z^2 + z^3 - z^4 + \cdots. \tag5 $$
Introduce the generating function of Catalan numbers
$$ c(x) := \sum_{n=0}^\infty C_n x^n = 1 + x + 2x^2 + 5x^3 + \cdots. \ \tag6 $$
The usual well-known recurrence for Catalan numbers is
$$ C_{n+1} = \sum_{k=0}^n C_k C_ {n-k} \tag7 $$
with many direct combinatorial interpretations. Expressed in terms of the generating function $\,c (x)\,$ this is equivalent to
$$ c(x) = 1 + x\,c (x)^2 = 1 + (c(x) x) c(x). \tag8 $$
This is also algebraically equivalent to
$$ (c(x)-1) = (c(x)x) + (c(x)x)(c(x)-1). \tag9 $$
Iterating this equality to the limit produces the infinite series equation
$$ (c(x)-1) = (c(x)x) + (c(x)x)^2 + (c(x)x)^3 + \cdots. \tag{10} $$
This has direct combinatorial interpretations also. For example, every Dyck word splits uniquely into nonempty irreducible Dyck words each of which uniquely corresponds to a Dyck word after removing the first and last letters.
Apply equation $(5)$ to this equation to get
$$ (c(x)x) = (c(x)-1) - (c(x)-1)^2 + (c(x)-1)^3 - \cdots. \tag{11} $$
The question is how to prove this combinatorially. A difficulty is how to interpret the alternating sum combinatorially. Perhaps one way is to use inclusion-exclusion?
It would be sufficient to find a combinatorial proof of equation $(5)$ where both $\,y\,$ and $\,z\,$ are formal power series in $\,x\,$ with constant term $0$. More explicitly,
$$ y = a_1x + a_2 x^2 + a_3 x^3 + \cdots \;\; \text{ and } \;\; z = b_1x + b_2 x^2 + b_3 x^3 + \cdots. \tag{12} $$
Use equation $(5)$ to get
$$ b_1 = a_1,\;\; b_2 = a_1^2 + a_2, \;\; b_3 = a_1^3 + 2a_1a_2 + a_3, \;\; \dots \tag{13} $$
whose combinatorial interpretation is a generalization of the one for equation $(10)$ with Dyck words. That is, $\,b_n\,$ represents all Dyck words of length $\,2n\,$ while $\,a_n\,$ represents all irreducible Dyck words of length $\,2n.\,$
Also use equation $(5)$ to get
$$ a_1 = b_1,\;\; a_2 = -b_1^2 + b_2, \;\; a_3 = b_1^3 - 2b_1b_2 + b_3, \;\; \dots \tag{14} $$
whose combinatorial interpretation would be a generalization of equation $(11)$ but is currently unknown to me. A combinatorial proof would be nice to know. Does anyone know?
As the question observes, each Dyck word decomposes uniquely as a concatenation of nontrivial irreducible Dyck words; say that a concatenation of $\ell$ irreducible Dyck words has $\ell - 1$ returns. $xc(x)$ is the gf for irreducible Dyck words. $c(x) - 1$ is the gf for nontrivial Dyck words. $(c(x) - 1)^k$ is the gf for concatenations of $k$ Dyck words. Therefore, each a Dyck word $D$ with exactly $m$ returns is counted in $(c(x) - 1)^k$ precisely $\binom{m}{k - 1}$ times. Taking the signs into account, this means $D$ is counted $\sum_k (-1)^k \binom{m}{k - 1}$ times on the right side of (11); this is $0$ if $m > 0$ and is $1$ if $m = 0$, i.e., if $D$ is irreducible.
(None of this has anything to do with the fact that $c(x)$ is the GF for Dyck words specifically; the same combinatorics governs the behavior of (4) and (5) in general, where $y$ is the gf for irreducibles and $z$ is the gf for concatenations of irreducibles.)