Background (may be skipped by those interested only in the basic question and not important associations):

“An essay on continued fractions” by Euler (translated by Myra and Bostwick Wyman) contains on page 319 Euler's continued fraction $$ \begin{split} q & = \frac{a}{p} + \frac{1}{3\dfrac{a}{p}+\dfrac{1 }{5\dfrac{a}{p}+\dfrac{1}{7\dfrac{a}{p}+\dfrac{ 1}{\ddots}} } \; \;}\\ &= CF\left[\;\frac{a}{p}, \; 3\frac{a}{p}, \; 5 \frac{a}{p}, ..., \; (2n+1)\frac{a}{p}, \; ... \;\right]. \end{split} $$

The CF for the reciprocal is

$$ \bar{q} = \frac{1}{q} = CF\left[0,\;\frac{a}{p}, \; 3\frac{a}{p}, \; 5 \frac{a}{p}, ..., \; (2n+1)\frac{a}{p}, \; ... \;\right].$$

Euler shows that the value of $q$ is defined by the Riccati equation

$$a \; dq + q^2 dp = dp.$$

$$ \frac{dq}{dp} = q'(p) = \frac{1-q^2}{a} .$$

Then from the formulas in my contribution (Sept 18, 2014) to OEIS 008292 on the Eulerian numbers with $\hat{p} = \frac{p}{a}$,

1)

$$ \frac{1}{q} = \bar{q} = \frac{e^{\hat{p}}-e^{-\hat{p}}}{e^{\hat{p}}+e^{-\hat{p}}} = \tanh(\hat{p}) =A(\hat{p}, \;1, \; -1),$$

where

$$A(x,a,b)= \frac{e^{ax}-e^{bx}}{a\;e^{bx}-b\;e^{ax}} = x + (a+b) \;\frac{x^2}{2!} + (a^2+4\;ab+b^2)\; \frac{x^3}{3!} + (a^3+11\;a^2b+11\;ab^2+b^3) \; \frac{x^4}{4!} + ...$$

is an e.g.f. for the bivariate Eulerian polynomials $E_n(a,b)$, whose coefficients are those of the h-vectors for the permutohedra,

2)

$$ \frac{p}{a} = \hat{p} = \frac{1}{2} \; \ln\left[ \;\frac{1+\bar{q}}{1-\bar{q}} \;\right] = \tanh^{(-1)}(\bar{q}) = B(\bar{q}, \;1, \; -1),$$

where

$$ B(x,a,b) = \frac{1}{a-b} \; \ln\left[ \; \frac{1+ax}{1+bx} \;\right] = x - (a+b) \; \frac{x^2}{2} + (a^2+ab+b^2)\; \frac{x^3}{3} - (a^3+a^2b+ab^2+b^3) \; \frac{x^4}{4} + ... $$

$$= \ln(1+u.\;x),$$

with $(u.)^n = u_n = h_{n-1}(a,b)$ a complete homogeneous polynomial in two indeterminates with $h_n(1,x)$ the h-vector of the ${(n-1)}$-dimensional hypertetrahedron, is a log generating function for the complete homogeneous polynomials,

3)

$$ \frac{d\bar{q}}{d\hat{p}} = \bar{q}' = (1+\bar{q})\;(1-\bar{q}),$$

an instance of the Riccati equation

$$ D_x \; A(x,a,b) = A'(x,a,b) = (1+a \;A)\;(1+b\;A),$$

which can be written in terms of a Weierstrass elliptic function (see Buchstaber & Bunkova in the OEIS entry)

4)

$$ \frac{1}{q} = \bar{q}(\hat{p}) = e^{\hat{p}\; (1-u^2) \; D_u} \; u \; |_{u=0},$$

more generally the bivariate Eulerian row polynomials $E_n(a,b)$ of $A(x,a,b)$ with $E_0(a,b) =0$ are generated by

$$ E_n(a,b) = [\;(1+ax) \; (1+bx) \; D_x\;]^{n} \; x \; |_{x=0}$$

(see OEIS A145271 for a generator of compositional inverses via the refined Eulerian numbers)

So, with $x = a/p$ the continued fraction of Euler evaluates analytically as

$$q(x) = \coth\left(\frac{1}{x}\right) = \coth\left(\frac{p}{a}\right)$$

with a discontinuity--a jump from $-1$ to $1$ as the argument passes through the origin from negative to positive values of $x$.

The reciprocal, of course, is

$$\bar{q} = \frac{1}{q} = \tanh\left(\frac{1}{x}\right) = \tanh\left(\frac{p}{a}\right)$$

with the same discontinuity at the origin $x=0$.

The more natural presentation is

$$\bar{q} = \tanh\left(\frac{p}{a}\right)$$

with no discontinuity for finite real argument $\frac{p}{a}$.

Note:

$$FGL(x,y) = A(B(x,a,b) + B(y,a,b),a,b) = \frac{x+y+(a+b)xy}{1-ab\; xy}$$

is called the hyperbolic formal group law and related to a generalized cohomology theory proposed by Lenart and Zainoulline.

For $a=-b=1$

$$FGL_{\tanh}(x,y) = \frac{x+y}{1+\; xy}$$

This is the addition, or composition, law for velocities in special relativity for $c=1$ and the formula for the hyperbolic tangent of sums

$$ \tanh(z+w) = \frac{\tanh(z)+\tanh(w)}{1+\tanh(z) \tanh(w)}.$$

See my post "The Elliptic Lie Triad: KdV and Riccati Equations, Infinigens, and Elliptic Genera" for relationships to a soliton solution to the KdV equation and an associated Riccati equation or my contribution to the MO-Q "Is there an underlying explanation for the magical powers of the Schwarzian?" for a briefer note on some aspects of the relationships.

Question:

What are continued fraction reps for

$$A(x,a,b)= \frac{e^{ax}-e^{bx}}{a\;e^{bx}-b\;e^{ax}} = x + (a+b) \;\frac{x^2}{2!} + (a^2+4\;ab+b^2)\; \frac{x^3}{3!} + (a^3+11\;a^2b+11\;ab^2+b^3) \; \frac{x^4}{4!} + ...$$

and what references for any specific rep are available (via the usual free sources)?

I suspect some version of Equation 4 in "Introduction to Chapter 3 on continued fractions [version 5, 29 January 2013]" by Xavier Viennot interpreted in terms of Dyck lattice paths should apply since the associahedra partition polynomials of OEIS A133437 for compositional inversion can be applied to $B(x,a,b)$ to obtain $A(x,a,b)$ and these associahedra face polynomials are a refinement of those of A126216, which are related to marked Dyck paths (and Schroeder--see Drake therein).

Another potential lead is A134264 / A125181 for compositional inversion via noncrossing partitions / Dyck paths of even length. See "A note on 2-distant noncrossing partitions and weighted Motzkin paths" by Ira Gessel and Jang Soo Kim, related to CFs.

EDIT 7/3/21:

I've scanned over dozens of references on orthogonal polynomials and continued fractions over the last couple of weeks, but didn't come across until just now "Lattice paths and branched continued fractions: An infinite sequence of generalizations of the Stieltjes–Rogers and Thron–Rogers polynomials, with coefficientwise Hankel-total positivity" by Mathias Petreolle, Alan Sokal, and Bao-Xuan Zhu. The footnote on p.77 states:

The identity (12.6) — that is, the S-fraction for the Eulerian polynomials — was found by Stieltjes [160, section 79]. Stieltjes does not specifically mention the Eulerian polynomials, but he does state that the continued fraction is the formal Laplace transform of $(1 − y)/(e^{t(y−1)} − y), which is well known to be the exponential generating function of the Eulerian polynomials. Stieltjes also refrains from showing the proof: “Pour abreger, je supprime toujours les artifices qu’il faut employer pour obtenir la transformation de l’int´egrale definie en fraction continue” (!). But a proof is sketched, albeit also without much explanation, in the book of Wall [165, pp. 207–208]. The J-fraction corresponding to the contraction of this S-fraction was proven, by combinatorial methods, by Flajolet [52, Theorem 3B(ii) with a slight typographical error]. Dumont [41, Propositions 2 and 7] gave a direct combinatorial proof of the S-fraction, based on an interpretation of the Eulerian polynomials in terms of “bipartite involutions of [2n]” and a bijection of these onto Dyck paths.

(The paper also contains, on p. 83, the partition polynomials of A190015, which they call the Euler symmetric polynomials, claiming to have introduced them originally in their paper. As noted in the OEIS entry, they are a scaled version of A145271, which I call the refined Eulerian partition polynomials mentioned above.)

Dumont's paper "Pics de cycle et derivees partielles" gives, on p. 38, the continued fraction for an o.g.f. of the bivariate Eulerian polynomials as

$$x + \sum_{n \geq 1} A_n(x,y)\; u^n $$

$$ \begin{split} & = \frac{x}{1-\dfrac{yu }{1-\dfrac{xu}{1-\dfrac{ 2yu}{\ddots}} } \; \;}\\ &= SFC[\; x, \; \frac{yu}{xu}, \; \frac{2yu}{2xu}, \; \frac{3yu}{3xu},..] \end{split} $$

so the question becomes what is the continued fraction for an e.g.f. given the continued fraction of its Laplace transform, or shifted o.g.f.?