$$u(x) = \int_{-\infty}^{\infty}f(s)G(x,s)ds = \int_{-\infty}^{\infty}f(x)G(x-s)ds$$ I was looking at an example where the integrand was written in the latter way and not the former and I wanted to know if this was simply a universal way of rewriting the integrand.
EDIT: The issue isn't so much why $G(x,s)$ is written by $G(x-s)$, the question more comes from changing $f(s)$ to $f(x)$. I don't know why this is the case.
Let us first fix notation here. A Green function for a linear differential operator $L$ is any solution of $$ L G(x, \xi) = \delta(x-\xi) $$ For example, if $L = \Delta$ is the Laplacian in 3 dimensions, then $$ G(x, \xi) = -\frac{1}{4 \pi |x-\xi|} $$ If a Green function $G$ is known, then a solution of $L u = f$ for general $f$ can be obtained by integration: $$ u(x) = \int_{\mathbb{R}^n} G(x, \xi) f(\xi) d\xi. $$
If the differential operator $L$ has constant coefficients (e.g. the Laplacian), then one can always find a Green function $G(x,\xi)$ which depends only on $x-\xi$. That is, one can find a function $E$ of one variable such that $ G(x, \xi) = E(x-\xi) $ and thus the integral above may be written $$ u(x) = \int_{\mathbb{R}^n} E(x-\xi) f(\xi) d\xi. $$ You may notice that this is precisely the convolution of $E$ and $f$, so you may rewrite this as $$ u(x) = (E * f)(x) = \int_{\mathbb{R}^n} E(x-\xi) f(\xi) d\xi = \int_{\mathbb{R}^n} E(\xi) f(x-\xi) d\xi. $$ Some authors conflate the two functions $G$ and $E$ and write $G(x, \xi) = G(x-\xi)$, while others consider this an abuse of notation.