An identity involving binomial coefficients
This is a nice example to apply the inversion rule of formal power series stated as rule 4 in G.P. Egorychevs Integral Representations and the Computation of Combinatorial Sums section 1.2.2.
I think it's worthwhile to present a complete proof. Here we use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a formal power series. We show
The following is valid for $m,n\geq 1$ and $a\in\mathbb{R}$ appropriate \begin{align*} \sum_{{i+j=m}\atop{i,j\geq 0}}&\frac{n-1}{ai+n-1}\binom{ai+n-1}{i}\frac{1}{aj+1}\binom{aj+1}{j} =\frac{n}{am+n}\binom{am+n}{m} \end{align*}
$$ $$
We obtain \begin{align*} \sum_{{i+j=m}\atop{i,j\geq 0}}&\frac{n-1}{ai+n-1}\binom{ai+n-1}{i}\frac{1}{aj+1}\binom{aj+1}{j}\\ &=\sum_{i=0}^m\frac{n-1}{ai+n-1}\binom{ai+n-1}{i}\frac{1}{am-ai+1}\binom{am-ai+1}{m-i}\\ &=\sum_{i=0}^\infty\left\{\binom{ai+n-1}{i}-a\binom{ai+n-2}{i-1}\right\}\\ &\qquad\qquad\cdot\left\{\binom{am-ai+1}{m-i}-a\binom{am-ai}{m-i-1}\right\}\tag{1}\\ &=\sum_{i=0}^{\infty}\left([z^i](1+z)^{ai+n-1}-a[z^{i-1}](1+z)^{ai+n-2}\right)\\ &\qquad\qquad\cdot\left([w^{m-i}](1+w)^{a(m-i)+1}-a[w^{m-i-1}](1+w)^{a(m-i)}\right)\\ &=[w^m](1+w)^{am}(1+w-aw)\\ &\qquad\qquad\cdot\sum_{i=0}^{\infty}\left(\frac{w}{(1+w)^a}\right)^i[z^i](1+z)^{n-2}(1+z-az)(1+z)^{ai}\tag{2}\\ &=[w^m](1+w)^{am}(1+w-aw)\left.\left((1+z)^{n-1}\right)\right|_{z=w}\tag{3}\\ &=[w^m](1+w)^{am+n}-a[w^{m-1}](1+w)^{am+n-1}\\ &=\binom{am+n}{m}-a\binom{am+n-1}{m-1}\\ &=\frac{n}{am+n}\binom{am+n}{m} \end{align*} and the claim follows.
Comment:
In (1) we use the identity $$\frac{q}{pk+q}\binom{pk+q}{k}=\binom{pk+q}{k}-p\binom{pk+q-1}{k-1}$$ We also set the upper limit of the sum to $\infty$ without changing anything since we are only adding zero.
In (2) we rearrange the sum, use the linearity of the coefficient of operator and the rule $[z^{n-k}]A(z)=[z^n]z^kA(z)$.
In (3) we use the inversion rule
\begin{align*} \sum_{i=0}^\infty w^i [z^i]A(z)f^i(z)=\left.\left(A(z)\frac{f(z)}{f(z)-zf^{\prime}(z)}\right)\right|_{z=g(w)} \end{align*} with $g(w)$ the inverse of $w=\frac{z}{g(z)}$. We get from (2) \begin{align*} A(z)&=(1+z)^{n-2}(1+z-az)\\ f(z)&=(1+z)^a \end{align*} and obtain \begin{align*} A(z)\frac{f(z)}{f(z)-zf^{\prime}(z)}&=(1+z)^{n-2}(1+z-az)\frac{(1+z)^a}{(1+z)^a-az(a+z)^{a-1}}\\ &=(1+z)^{n-1} \end{align*} Since $w=\frac{z}{f(z)}=\frac{z}{(1+z)^a}$ we apply in (3) the substitution $z=w$.