A conceptual way to tackle this question is to look at universal distributions on $\mathbf{Q}/\mathbf{Z}$, studied by Kubert and Lang among others. Distributions arise naturally in number theory, see e.g. Lang's article Relations de distributions et exemples classiques.
There is a function $f$ on $\mathbf{Q}/\mathbf{Z}$ with values in an abelian group $A$ which satisfies the distribution relations $\sum_{a \in \mathbf{Z}/N\mathbf{Z}} f(x+a/N)=f(Nx)$ for every $x \in \mathbf{Q}/\mathbf{Z}$ and every $N \geq 1$, and which is universal for this property. The construction of $f$ and $A$ is obvious: take the free abelian group on $\mathbf{Q}/\mathbf{Z}$ and mod out by the distribution relations.
Similarly, you can look at functions on $\frac{1}{N}\mathbf{Z}/\mathbf{Z}$ satisfying the $M$-distribution relations for every $M$ dividing $N$. You get an abelian group $A_N$ and a canonical map $A_N \to A$.
Kubert has shown in the article The universal ordinary distribution that $A$ is a free abelian group, that the map $A_N \to A$ is injective, that $A_N$ is free of rank $\varphi(N)$, and has determined explicit free generators of $A_N$.
The $\Gamma$ function determines a distribution with values in $\mathbf{C}^\times/\overline{\mathbf{Q}}^\times$. This distribution is odd and the Rohrlich-Lang conjecture asserts that it is universal, in other words the induced map $\overline{\Gamma} : A_N^- \to \mathbf{C}^\times/\overline{\mathbf{Q}}^\times$ should be injective, where $A_N^-$ denotes the odd part of $A_N$.
Concretely, you can write your purported relation as a formal $\mathbf{Z}$-linear combination of the $a/N$ with $0 < a <N$ and check whether this divisor belongs to the subspace generated by the $M$-distribution relations for every $M|N$ and by the relations $[x]+[-x]=0$. This is just linear algebra in a $\mathbf{Q}$-vector space of dimension $N$ so should be doable for quite large $N$.
EDIT. In fact, you can simply use a known universal distribution and evaluate it on your divisor. For example, the Bernoulli distribution $$\tilde{B}_1(x) = \sum_{u \in (\mathbf{Z}/N\mathbf{Z})^\times} B_1(ux) [u]$$ taking values in $\mathbf{Q}[(\mathbf{Z}/N\mathbf{Z})^\times]$ is an odd universal distribution. Here $B_1(y) = \{y\}-\frac12$ for every $y \in \mathbf{R}/\mathbf{Z}$ is the Bernoulli polynomial. So you can simply check whether your divisor is in the kernel of $\tilde{B}_1$.