As @ParamanandSingh commented, Dougall's original paper is linked here. Since the paper is written in an old style, it is quite hard to read. This answer is a modification of Dougall's original proof. Considering that currently there are no proofs typed on the internet, I will present the proof as detailed as possible so various level of math learners could understand.
I will denote the rising and falling factorials as
$$a^{\overline n}=a(a+1)\cdots(a+n-1) \quad ,\quad a^{\underline n}=a(a-1)\cdots(a-n+1)$$
Start by considering the function
$$f(x,s)=\frac{(x+a+1)^{\overline s}(x+b+1)^{\overline s}(c+1)^{\overline s}(d+1)^{\overline s}}{(a+1)^{\overline s}(b+1)^{\overline s}(x+c+1)^{\overline s}(x+d+1)^{\overline s}}$$
where $s$ is a positive integer and constrain the expression to $a+b=c+d$. We can see the following properties of $f$
- $f$ is symmetric ($f(x,s)=f(s,x)$)
- $f(x,s)$ is invariant under $x\mapsto -x-c-d-s-1$
Dougall then expand the function $f$ as an infinite sum, in a way such that the expansion posses the above two properties.
$$f(x,s)=\sum_{k=0}^{\infty}A_k \frac{x^{\underline k}s^{\underline k} (x+c+d+s+1)^{\overline k}}{(x+c+1)^{\overline k}(s+c+1)^{\overline k}(x+d+s)^{\underline k}}$$
where $A_k$ are coefficients independent of $s$ and $x$ yet determined. To find $A_k$, we multiply both sides by $(x+d+1)^{\overline s}$ and take the limit as $x\to -d-1$
$$\begin{split}\lim_{x\to -d-1}\{(x+d+1)^{\overline s}f(x,s)\}=\sum_{k=0}^{\infty}A_k \lim_{x\to -d-1}\left\{\frac{x^{\underline k}s^{\underline k} (x+c+d+s+1)^{\overline k}(x+d+1)^{\overline s}}{(x+c+1)^{\overline k}(s+c+1)^{\overline k}(x+d+s)^{\underline k}}\right\}
\end{split}$$
We note that since there is a $s^{\underline k}$ in the summand, the sum is null for $k>s$. now take a look at the expression
$$\frac{(x+d+1)^{\overline s}}{(x+d+s)^{\underline k}}=\begin{cases}1,\quad s=k \\ (x+d+s-k)\cdots(x+d+1), \quad k<s\end{cases}$$
and notice that if we take the limit $x\to -d-1$ the expression is non zero only if $k=s$, so the limit reduces to
$$\lim_{x\to -d-1}\{(x+d+1)^{\overline s}f(x,s)\}=A_s\lim_{x\to -d-1}\frac{x^{\underline s} s!(x+c+d+s+1)^{\overline s}}{(x+c+1)^{\overline s}(s+c+1)^{\overline s}}$$
$$=A_s\frac{(-d-1)^{\underline s}s!(s+c)^{\overline s}}{(c-d)^{\overline s}(s+c+1)^{\overline s}}=A_s\frac{(-1)^s(d+1)^{\overline s}s!(s+c)}{(c-d)^{\overline s}(2s+c)}$$
Where using the definition of $f$, the left hand side is equal to $$\lim_{x\to -d-1}\{(x+d+1)^{\overline s}f(x,s)\}=\frac{(c+1)^{\overline s}(d+1)^{\overline s}}{(a+1)^{\overline s}(b+1)^{\overline s}}\frac{(a-d)^{\overline s}(b-d)^{\overline s}}{(c-d)^{\overline s}}$$
Putting together we get
$$A_n=\frac{(-1)^n}{n!}\cdot\frac{2n+c}{n+c}\cdot \frac{(a-d)^{\overline n}(b-d)^{\overline n}(c+1)^{\overline n}}{(a+1)^{\overline n}(b+1)^{\overline n}}$$
To see symmetry, we write $a$ as $y+c$, $b$ as $z+c$ and $d$ as $z+y+c$ to change the expression into the form
$$\frac{(x+y+c+1)^{\overline s}(z+y+c+1)^{\overline s}(x+z+c+1)^{\overline s}(c+1)^{\overline s}}{(x+c+1)^{\overline s}(y+c+1)^{\overline s}(z+c+1)^{\overline s}(x+y+z+c+1)^{\overline s}}$$
$$=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}\frac{2k+c}{k+c}\frac{(-z)^{\overline k}(-y)^{\overline k}(c+1)^{\overline k}}{(y+c+1)^{\overline k}(z+c+1)^{\overline k}}\cdot \frac{x^{\underline k}s^{\underline k}(x+y+z+s+2c+1)^{\overline k}}{(x+c+1)^{\overline k}(s+c+1)^{\overline k}(x+y+z+c+1)^{\underline k}}$$
Finally, we take $s$ to infinity. Note that $\lim_{s\to\infty}\frac{\Gamma(s+a)}{\Gamma(s)a^s}=1$. The left hand side is equal to
$$\frac{\Gamma(x+c+1)\Gamma(y+c+1)\Gamma(z+c+1)\Gamma(x+y+z+c+1)}{\Gamma(c+1)\Gamma(x+y+c+1)\Gamma(x+z+c+1)\Gamma(y+z+c+1)}$$
The right hand side is equal to
$$=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}\frac{2k+c}{k+c}\frac{(-z)^{\overline k}(-y)^{\overline k}(c+1)^{\overline k}x^{\underline k}}{(y+c+1)^{\overline k}(z+c+1)^{\overline k}(x+c+1)^{\overline k}}\cdot {\underbrace{\lim_{s\to\infty}\frac{s^{\underline k}(x+y+z+s+2c+1)^{\overline k}}{(s+c+1)^{\overline k}(x+y+z+c+1)^{\underline k}}}_{=1}}$$
$$=\sum_{k=0}^{\infty}\frac{2k+c}{c}\cdot \frac{(-x)^{\overline k}(-y)^{\overline k}(-z)^{\overline k}c^{\overline k}}{(x+c+1)^{\overline k}(y+c+1)^{\overline k}(z+c+1)^{\overline k}k!}$$
$$={}_5F_4\left(\begin{array}{cc}c/2+1, c ,-x,-y,-z \\
c/2 ,x+n+1,y+n+1,z+n+1\end{array};1\right)$$
This completes the proof.
As you can see, the proof utilizes no advanced theory of Hypergeometric series. All of it were elementary algebraic manipulations and a clever expansion.
Also, note that the condition writing as an infinite sum is $\Re(x+y+z+c)>-1$.
PS. If we take $z\to\infty$, we get
$$\sum_{n=0}^{\infty}\frac{2n+c}{c}\frac{(-x)^{\overline n}(-y)^{\overline n}c^{\overline n}}{(x+c+1)^{\overline n}(y+c+1)^{\overline n}n!}(-1)^n =\frac{\Gamma(x+c+1)\Gamma(y+c+1)}{\Gamma(c+1)\Gamma(x+y+c+1)}$$
For example, set $c=1/2$, $x=y=-1/2$ gives
$$\frac{2}{\pi}=\sum_{n=0}^{\infty}\frac{(4n+1)}{4^{3n}}\binom{2n}{n}^3(-1)^n$$
