Found this identity by accident

There may be a more specific name for this formulae, but it follows almost immediately from Gauss' Formulae: $$_2F_1(a,b;c;1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$$ where $_2F_1$ denotes the Hypergeometric function. To see this, note that we have that by the power-series definition for $_2F_1$, we have that: $$\lim_{q\to 0}\frac{1}{q}(_2F_1(q,1;x+1;1)-1)=\sum_{i=1}^{\infty}\frac{(n-1)!}{\prod_{k=1}^{n}(x+k)}$$ On the other hand we have that: $$\frac{\Gamma(x+1)\Gamma(x-q)}{\Gamma(x)\Gamma(x+1-q)}=\frac{x}{x-q}$$ and it is clear that: $$\lim_{q\to 0}\frac{1}{q}(\frac{x}{x-q}-1)=\frac{1}{x}$$ Putting the sides together you get your identity.

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\begin{align} &\bbox[5px,#ffd]{\left.\sum_{n = 1}^{\infty}{\pars{n - 1}!\,x \over \prod_{k = 1}^{n}\pars{k + x}} \,\right\vert_{\ x\ >\ 0}} = x\sum_{n = 1}^{\infty}{\Gamma\pars{n} \over \pars{1 + x}^{\large\overline{n}}} \\[5mm] = &\ x\sum_{n = 1}^{\infty}{\Gamma\pars{n} \over \Gamma\pars{1 + x + n}/\Gamma\pars{1 + x}} = x\sum_{n = 1}^{\infty}{\Gamma\pars{n}\Gamma\pars{x + 1}\over \Gamma\pars{n + x + 1}} \\[5mm] = &\ x\sum_{n = 1}^{\infty} \int_{0}^{1}t^{n - 1}\pars{1 - t}^{x}\dd t = x\int_{0}^{1}{1 \over 1 - t}\pars{1 - t}^{x}\dd t \\[5mm] = &\ x\int_{0}^{1}t^{x - 1}\,\dd t = \bbx{\large 1} \\ & \end{align}