We first use the differentiation formula for the generalized hypergeometric function
\begin{equation}
\frac{a_1a_2\dots a_{p}}{b_1b_2\dots b_q}{}_pF_q\left(\left.\begin{array}{c} c_1,c_2,\dots ,c_p\\ d_1,d_2,\dots ,d_q \end{array}\right| z\right)=\frac{d}{dz}{}_pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q \end{array}\right| z\right)
\end{equation}
Then, the LHS of the proposed identity can be written as
\begin{equation}
_pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q \end{array}\right| z\right)+z\,\frac{a_1a_2\dots a_{p-1}}{b_1b_2\dots b_q}{}_pF_q\left(\left.\begin{array}{c} c_1,c_2,\dots ,c_p\\ d_1,d_2,\dots ,d_q \end{array}\right| z\right)=\left( 1+\frac{z}{a_p}\frac{d}{dz} \right){} _pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q \end{array}\right| z\right)\tag{1}\label{eq1}
\end{equation}
To differentiate the hypergeometric function, we use the Euler's integral transform
\begin{align}
& _pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q \end{array}\right| z\right)\\
&=\frac{\Gamma(b_q)}{\Gamma(a_p)\Gamma(b_q-b_p)} \int_0^1t^{a_p-1}\left( 1-t \right)^{b_q-a_p-1}{}_{p-1}F_{q-1}\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_{p-1}\\ b_1,b_2,\dots ,b_{q-1} \end{array}\right| t\right)\,dt
\end{align}
Here $b_q=a_p+1$, then
\begin{align}
_pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q \end{array}\right| z\right)&=
a_p \int_0^1t^{a_p-1}{}_{p-1}F_{q-1}\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_{p-1}\\ b_1,b_2,\dots ,b_{q-1} \end{array}\right| zt\right)\,dt\\
&=\frac{a_p}{z^{a_p}} \int_0^zu^{a_p-1}{}_{p-1}F_{q-1}\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_{p-1}\\ b_1,b_2,\dots ,b_{q-1} \end{array}\right| u\right)\,du
\end{align}
Then
\begin{align}
\frac{d}{dz}&\,{} _pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q \end{array}\right| z\right)\\
&=\frac{a_p}{z}\,{}_{p-1}F_{q-1}\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_{p-1}\\ b_1,b_2,\dots ,b_{q-1} \end{array}\right| z\right)-\frac{a_p}{z} \,{}_pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q \end{array}\right| z\right)
\end{align}
Plugging this expression in eq. \eqref{eq1} we find theRHS of the proposed identity.
Simplify the Pochhammer symbols with the identity $(a)_n=\dfrac{\Gamma(a+n)}{\Gamma(a)}$ and you'll end up with zeta-related sums:
$$\begin{align*}
{}_4F_3\left(\left.\begin{array}{c}2,\frac32,\frac32,\frac32\\\frac52,\frac52,\frac52\end{array} \right\vert 1\right) &= \sum_{n=0}^\infty \frac{(2)_n \left[\left(\frac32\right)_n\right]^3}{n! \left[\left(\frac52\right)_n\right]^3} \\
&= \sum_{n=0}^\infty \frac{\frac{\Gamma(n+2)}{\Gamma(2)} \left(\frac{\Gamma\left(n+\frac32\right)}{\Gamma\left(\frac32\right)}\right)^3}{\Gamma(n+1) \left(\frac{\Gamma\left(n+\frac52\right)}{\Gamma\left(\frac52\right)}\right)^3} \\
&= \left(\frac32\right)^3 \sum_{n=0}^\infty \frac{n+1 }{\left(n+\frac32\right)^3} \\
&= 27 \sum_{n=1}^\infty \frac{n}{(2n+1)^3} \\
&= \frac{27}2 \sum_{n=1}^\infty \left(\frac1{(2n+1)^2} - \frac1{(2n+1)^3}\right)
\end{align*}$$
Now observe
$$\begin{align*}
\zeta(a) &= \sum_{n=1}^\infty \frac1{n^a} \\
&= 1 + \sum_{n=1}^\infty \frac1{(2n)^a} + \sum_{n=1}^\infty \frac1{(2n+1)^a} \\[1ex]
\implies \sum_{n=1}^\infty \frac1{(2n+1)^a} &= \left(1-2^{-a}\right)\zeta(a)-1
\end{align*}$$
Best Answer
$$S=\sum_{k\geq0}\frac{(a_1)_k\cdots(a_{p})_k}{(b_1)_k\cdots(b_{q})_k}\frac{kx^k}{k!}$$ $$S=\frac{(a_1)_0\cdots(a_{p})_0}{(b_1)_0\cdots(b_{q})_0}\frac{0x^0}{0!}+\sum_{k\geq1}\frac{(a_1)_k\cdots(a_{p})_k}{(b_1)_k\cdots(b_{q})_k}\frac{kx^k}{k!}$$ $$S=\sum_{k\geq1}\frac{(a_1)_k\cdots(a_{p})_k}{(b_1)_k\cdots(b_{q})_k}\frac{kx^k}{k!}$$ $$S=x\sum_{k\geq1}\frac{(a_1)_k\cdots(a_{p})_k}{(b_1)_k\cdots(b_{q})_k}\frac{x^{k-1}}{(k-1)!}$$ $$S=x\sum_{k\geq0}\frac{(a_1)_{k+1}\cdots(a_{p})_{k+1}}{(b_1)_{k+1}\cdots(b_{q})_{k+1}}\frac{x^{k}}{k!}$$ Note that $(a)_{n+1}=a\cdot(a+1)_n$. Thus, $$S=x\sum_{k\geq0}\frac{a_1\cdots a_{p}}{b_1\cdots b_{q}}\frac{(a_1+1)_k\cdots(a_{p}+1)_k}{(b_1+1)_k\cdots(b_{q}+1)_k}\frac{x^{k}}{k!}$$ $$S=x\frac{a_1\cdots a_{p}}{b_1\cdots b_{q}}\sum_{k\geq0}\frac{(a_1+1)_k\cdots(a_{p}+1)_k}{(b_1+1)_k\cdots(b_{q}+1)_k}\frac{x^{k}}{k!}$$ $$S=x\frac{a_1\cdots a_{p}}{b_1\cdots b_{q}}\;_pF_q\big(a_1+1,\dots,a_{p}+1;b_1+1,\dots,b_q+1;x\big)$$ And there you go.