To avoid all poles in the Mellin inversion formula you want to integrate along the line $\int_{\gamma-i\infty}^{\gamma+i\infty}ds$ where $\gamma>\max(0,-a/2,-b/2)$; then Mathematica gives the result in terms of the <A HREF="https://en.wikipedia.org/wiki/Meijer_G-function">Meijer G-function</A>, see also Bateman and Erdélyi, Higher Transcendental Functions, Vol. I, paragraph 5.3.1:

$$F\left ( x \right )=\frac{1}{2i\pi}\int_{\gamma-i\infty}^{\gamma+i\infty} \Gamma(s)\Gamma (2s+a)\Gamma( 2s+b)x^{-s}\,ds=\frac{1}{\pi}2^{a+b-2} G_{0,5}^{5,0}\left(\frac{x}{16}|
\begin{array}{c}
 0,\frac{a}{2},\frac{a+1}{2},\frac{b}{2},\frac{b+1}{2} \\
\end{array}
\right).$$

The integral can also be evaluated by contour integration, closing the contour in the left-half complex plane and picking up the poles at $-n$, $-(n+a)/2$, $-(n+b)/2$, $n=0,1,2,\ldots$. For this we assume that $a\neq b$ and $a,b$ are both non-integer --- so that all poles are simple. I then find that

\begin{align}
F(x)=&\sum_{n=0}^{\infty }\frac{\left ( -1 \right )^{n}}{n!}\Gamma \left ( a-2n \right )\Gamma \left ( -2n+b \right )x^{n}\\
&+\frac{1}{2}\sum_{n=0}^{\infty}\frac{\left ( -1 \right )^{n}}{n!}\Gamma \left ( -\frac{n}{2}-\frac{a}{2} \right)\Gamma \left ( -n-a+b\right )x^{\frac{n}{2}+\frac{a}{2}}\\
&+\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\Gamma \left (-\frac{n}{2}-\frac{b}{2} \right )\Gamma (-n-b+a)x^{\frac{n}{2}+\frac{b}{2}}.
\end{align}
I checked numerically that this agrees with the Meijer G-function result -- it differs from the result in the OP by factors 1/2 in the second and third sum.