Well I'm not certain, but your operators seem to be like a modified lower hyperoperators. These operators satisfy the recursion:
\( (x[s]y)[s-1] x = x[s](y+1) \)
where \( x[0]y = x+y \)
Check if this function works. I'm pulling it out of a hat but I have a lot of math behind it.:
\( \frac{2}{x[s]y} = \frac{1}{\Gamma(-s)} \sum_{n=0}^\infty \frac{(-1)^n}{(x[n]y)n!(n-s)} + \sum_{k=0}^\infty \frac{a_k}{\Gamma(k-s+1)} \)
where
\( a_k = \sum_{n=0}^\infty \frac{(-1)^k}{x[n+k]y} \)
This may or may not work. Depending on if 1/x[s]y is holomorphic or meromorphic and if (x[s]y)[s-1]x is holo as well with decent enough behaviour.
\( (x[s]y)[s-1] x = x[s](y+1) \)
where \( x[0]y = x+y \)
Check if this function works. I'm pulling it out of a hat but I have a lot of math behind it.:
\( \frac{2}{x[s]y} = \frac{1}{\Gamma(-s)} \sum_{n=0}^\infty \frac{(-1)^n}{(x[n]y)n!(n-s)} + \sum_{k=0}^\infty \frac{a_k}{\Gamma(k-s+1)} \)
where
\( a_k = \sum_{n=0}^\infty \frac{(-1)^k}{x[n+k]y} \)
This may or may not work. Depending on if 1/x[s]y is holomorphic or meromorphic and if (x[s]y)[s-1]x is holo as well with decent enough behaviour.