How to prove that $$\sum _{j=0}^{n-i} \frac{(-1)^j \binom{n-i}{j}} {(i+j) (i+j+1) (n+i+j+1)\binom{n+i+j}{n-i}} =\frac{4 (2 i-1)!\, (2 n-2 i+1)!}{(2n+2)!},$$ where $n$ and $i$ are integers such $1\le i\le n$?
Mathematica knows this identity, which is also confirmed for $n=1,\dots,10$ and $i=1,\dots,n$.
The above identity is was the missing part in this proof.
In terms of hypergeometric series, the sum is $$ \frac{(2i)!\,(n-i)!}{i(i+1)(n+i+1)!}\,{}_3F_2\left({2i+1,\atop}{i,\atop i+2,}{-n+i\atop n+i+2}\biggm| 1\right) $$ which can be evaluated by Dixon's Theorem.
I came late to the game here, but let me add another variation for the broader audience. The tool is called the Wilf-Zeilberger method. It is an automatic process. It goes as follows. For convenience, replace $n\rightarrow n-i$.
Define the two functions \begin{align} F(n,j)&:=\frac{(-1)^j \binom{n}j(i + n + 1)(2n + 2i + 1) \binom{2n + 2i}{2n + 1}}{2(i + j)(i + j + 1)(n + 2i + j + 1) \binom{n + 2i + j}n}, \\ G(n,j)&:= - \frac{F(n,j)\,j(i+j+1)}{(2n+3)(n+1-j)}. \end{align} Next, verify that $F(n+1,j)-F(n,j)=G(n,j+1)-G(n,j)$. Now, sum both sides over all integers $j$ (understanding that $\binom{a}b=0$ if $a<b$, hence we have finite sums). The novel idea is the RHS vanishes while $f(n):=\sum_j F(n,j)$ is your proposed identity (after divided by its RHS). So, we are hence obtaining the relation $f(n+1)=f(n)$. Finally, just check that $f(1)=1$. This complete the argument.
