Can I split $\frac{1}{a-b}$ into the form $f(a)+f(b)$?

$f(a)+f(b)$ is symmetric: it does not change if you switch $a$ and $b$. But $\frac 1 {a-b}$ is not symmetric. So you cannot do this.

In fact, we can't even split it as $$\frac{1}{a-b}=f(a)+g(b)$$ for any functions $f,g$.

Suppose instead that functions $f,g$ from $\mathbb{R}$ to $\mathbb{R}$ are such that $$f(a)+g(b)=\frac{1}{a-b}$$ for all $a,b\in\mathbb{R}$ with $a\ne b$.

Then we would have \begin{align*} & \begin{cases} f(x+1)+g(x)={\Large{\frac{1}{(x+1)-x}}}={\Large{\frac{1}{1}}}=1\\[4pt] f(x-1)+g(x)={\Large{\frac{1}{(x-1)-x}}}={\Large{\frac{1}{-1}}}=-1\\ \end{cases} \\[6pt] \implies\!\!\!\!&\;\;\;\;f(x+1)-f(x-1)=2 \qquad\qquad\qquad\qquad\qquad \\[4pt] \end{align*} but we would also have \begin{align*} & \begin{cases} f(x+1)+g(x-2)={\Large{\frac{1}{(x+1)-(x-2)}}}={\Large{\frac{1}{3}}}\\[4pt] f(x-1)+g(x-2)={\Large{\frac{1}{(x-1)-(x-2)}}}={\Large{\frac{1}{1}}}=1\\ \end{cases} \\[6pt] \implies\!\!\!\!&\;\;\;\;f(x+1)-f(x-1)=\frac{1}{3}-1=-\frac{2}{3}\\[4pt] \end{align*} contradiction.