Convolution theorem for "weighted" convolution?
1) The terms autoconvolution or selfconvolution can be found in the literature, without the kernel $s$. So you would probably have a weighed, kernel or windowed self-convolution.
However, i foresee two potential caveats on your formulation. First, i do not understand why you require a two-dimensional periodic signal (and nothing on the kernel $s$?). This could result in infinite values of your integrals. Second, don't you assume some kind of symmetry in $s(r,r′)$?

2) Suppose you can solve your system with $s=1$, then with appropriate symmetry on $s$ you may be able to factorize $s$ and slit it over $g(r')$ and $g(r-r')$. Unfortunately the convolution theorem is likely to give you a family of solutions $g$ knowing $f$ only.

3) You may find useful to explore a multiscale form of your weighted self-correlation operator.

Affine invariant pattern recognition using Multiscale Autoconvolution
Rahtu E., Salo M., Heikkilä J.
IEEE Trans Pattern Anal Mach Intell. 2005 Jun;27(6):908-18.

An iterative deautoconvolution algorithm for nonnegative functions
Kerkil Choi and Aaron D Lanterman 
2005 Inverse Problems 21 981

Autoconvolution and panorama: Augmenting second-order signal analysis 
Douglas, S.C. and  Mandic, D.P., 2014 IEEE International Conference on Acoustics, Speech and Signal Processing (ICASSP)