A quick annotation:
I'm always sceptic when a weird function is spitted out by any mathematics program. With this in mind, I took the liberty to do the following analysis:
Suppose $a \not \in \mathbb{Z}$, and $\beta \neq 0$. Using Euler's reflection formula
$$
\Gamma(a) \Gamma(1-a) = \frac{\pi}{\sin \pi a}
$$
is easy to see that
$$
\beta^{-a} \Gamma(a) \sin(a \pi) +
e^\beta \beta^{2 a - 1} \Gamma(1 - a) \sin(a \pi) = 0
$$
transforms into
$$
1 + \frac{\pi}{\Gamma^2(a)\sin (\pi a)}e^\beta \beta^{3a-1} = 0.
$$
Then, for $\beta > 0$, there are solutions when $(2n - 1) < a < 2n$, for $n \in \mathbb{Z}$. Now, what happens when $a \rightarrow n$?
If $-1 < n$ then
$$
\lim_{a \rightarrow n} \bigl(\beta^{-a} \Gamma(a) \sin(a \pi) +
e^\beta \beta^{2 a - 1} \Gamma(1 - a) \sin(a \pi)\bigl) = \frac{\pi}{(n-1)!} \beta^{2n-1}e^\beta,
$$
so $\beta$ should be equal to zero, but then, in order to apply the implicit function theorem, the limits in the reverse order should exist, which is not the case.
If $n < -1$ the analog argument applies.
Now, if $\beta < 0$, the condition that $a \in \mathbb{R}$ implies that there is no solution at all.
Whit all this in mind, the implicit function theorem will guaranty that $a =a(\beta)$ as long as
$$
\partial_a \left\{\frac{\pi}{\Gamma^2(a)\sin (\pi a)}e^\beta \beta^{3a-1} \right\}\neq 0
$$
and $\beta = \beta(a)$ if we take $\partial_\beta\{\}$ instead. With a bit more of effort, one can show that the ContourPlot
shown by Mark is qualitatively correct (not that it wasn't in the first place, but since the original expression is so obscure, it's hard to see -at least for me- if there aren't any artifacts behind the plot), and continue to work with confidence.