This answer is an attempt to explain lemma1. Why is $S_1(z)$ the correct initial approximation in the previous answer, and why does the sequence converge?
$$S_1(z)=L-\lambda^z=L-e^{z \ln(\ln(L))}\;\;\;S_n(z)=\log_a(S_{n-1}(z+1))$$

Starting with ideas from the previous answer, along with the definition of the inverse Schröder function, which we will call $h(z)$, where the desired $S(z)$ function follows directly from $h(z)$ as follows: $S(z)=\sum a_n(-1)^n\lambda^{n z}\;\;$ where $a_n$ are the Taylor series coefficients of $h(z)$. We start with the fixed point L from the previous answer. $\ln_a(L+z) = L +\frac{z}{\lambda} + O z^2\; $ and the multiplier at the fixed point is $\lambda$.
$$L= \frac{W(-\ln a)}{-\ln a}\;\;\;\lambda=\ln(L) = \ln(a) \cdot L; \;\:1<L<e;\;\;0<\lambda<1$$

$$h(z)=\sum_{n=0}^{\infty}a_n z^n\;\;\;a_0=L;\;a_1=1\;\;\;h(\lambda z) = a^{h(z)}\;\;\;\ln_a(h(\lambda z))= h(z)$$
$$S(z) = h(\lambda^z)\;\;\;S(z+1)=a^{S(z)}\;\;\;\lim_{\Re(z) \to \infty}S(z)=L$$

**lemma4:** if all of the odd Taylor series coefficients of $h(z)$ at z=0 are positive, and all of the even Taylor series coefficients are negative, then $S(z)$ is fully monotonic, since $0<\lambda<1$, and $\lambda^{nz}=-\exp(n\ln(\lambda) z)$ is fully monotonic for all n, since $\ln(\lambda)$ is a negative number, and $a_n(-1)^n$ is also a negative number, so $S(z)$ is fully monotonic under that case.

There is a formal solution for $h(z)$, as the inverse of the Schröder equation. But alternatively, we start with $h_1(z)=L+z$ and iterate as follows.

$$h_{m+1}(z)=\ln_a(h_m(\lambda z))\;\;\; h(z)=\lim_{m\to\infty}h_m(z)\;\;\;h(z)= \ln_a(h(\lambda z))$$

$$h_2(z)=\ln_a(h_1(z)) =\ln_a(L+\lambda z) $$
$$\ln_a(L+z) =\ln_a(L(1+z/L)) = L + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}z^n}{n \cdot \ln(a) \cdot L^n}$$
$$h_2(z)= \ln_a(L+\lambda z) = L + \sum_{n=1}^{\infty}\frac{(-1)^{n+1} \cdot \lambda \cdot z^n}{n \cdot \ln(a) \cdot L^n} $$
$$h_2(z)= \ln_a(L+\lambda z) = L + \sum_{n=1}^{\infty}\frac{(-1)^{n+1} \cdot z^n}{n \cdot L^{n-1}} \;\;\; a_0=L; a_1=1$$

Notice that $\ln_a(z)$ has all odd Taylor series coefficients positive, and all even Taylor series coefficients negative, just as desired. Notice that $h_2(z)$ has all odd Taylor series coefficients positive, and all even Taylor series coefficients negative, just as desired. Similar to lemma2, the composition $h_3(z)=\ln_a \circ h_2(\lambda z)$ has the same property, so $h_3(z)$ has all odd Taylor series coefficients positive, and all even Taylor series coefficients negative, just as desired. And by induction, the function $h_m(z)$ has all odd Taylor series coefficients positive, and all even Taylor series coefficients negative, just as desired, so this is also true of $h(z)$. **lemma5:** the $h_m(z)$ sequence converges, which has been shown in the literature of complex dynamics. Also, notice that iterating from $h_{m+1}$ from $h_m$ that $a_0=L$, and $a_1=1$, for the Taylor series of each version of $h_m$. This approach is identical but a little bit more rigorous than the iterations in the previous answer, since it directly shows the stability in iterating $h_n(z)$, which makes proving convergence of lemma5 easier. Let us say the golden values of $a_n$ for the formal inverse Schröder function are $g_n$, and we are doing m iterations to generate $h_m(z)$. Then by observation, $\forall n>2, \;a_n=g_n+O\lambda^{m+n}\;$ after m iterations. Formalizing this error estimate would be a way of showing that the $h_m(z)$ sequence converges to $h(z)$, and that lemma5 is valid.

So if lemma2, lemma4, and lemma5 hold, the conclusion is that $S(z)$ is fully monotonic at the real axis in its range of analyticity, where $\Re(z+1)>S^{-1}(0)$

Finally, as in the previous answer we generate the desired sexp function.
$$\text{sexp}_a(z)=S(z+k)\;\;\;k=S^{-1}(0)+1\;\;\;\text{sexp}_a(z)=a^{\text{sexp}(z-1)}$$

$\text{sexp}_a(z)$ is completely monotonic at the real axis if z>-2. Then the sequence $^n a=\text{sexp}_a(n)$ is also a completely monotonic sequence by lemma3 that if an analytic function is completely monotonic over a given range, then a sequence of equally spaced samples of that function in that range is also completely monotonic.