Number of walks from $(0,0,\dots, 0)$ to $(1,1,\dots, 1)$ along the $N$-cube's edges is enumerated by $\sinh^N$ and loops from a vertex to itself by $\cosh^N$ so why isn't the first arrivals to the opposite corner enumerated by
$$\frac{\sinh^N(z)}{\cosh^N(z)}?$$
An arrival is a first arrival and a loop, isn't it?
Or am I wrong in that the product of first arrivals and loops isn't the star product of labelled combinatorial classes?
For context see for example Balls switching chambers: the Ehrenfest model. It is from this book (II.11. p. 118).
It is not the labeled star product. This would only be the case if you could arbitrarily choose which steps of the walk belonged to the "first arrival," and which belonged to the "loop." But every step in the former must come before the latter. You do not have a binomial convolution, just a convolution.
If you had the ordinary generating function $A(x)$ for number of walks to the opposite corner, and the OGF $B(x)$ for loops, then the OGF $C(x)$ for first arrivals to the opposite corner would be $A(x)/B(x)$. But it seems like $\cosh^N$ and $\sinh^N$ are EGFs.