The answer is no. The smallest counter-example: consider $\Gamma = K_3$, ans $\Gamma_1$ induced by a pair of vertices. In this case $G = S_3$ and has size 6, $G_0$ is trivial, and $|G_1| = 2$, while for the semiderict product we must have $|N \rtimes H| = |N| \times |H|$.

(Sorry for missing the "$G$ stabilizes $\Gamma_1$" condition in the previous edit)

The answer is no. Consider $\Gamma = ([4], \{12, 13, 23, 14\})$. $G$ stabilizes $\Gamma_1$ = the edge $14$. $G_0 = G$, and $|G_1| = 2$, while for the semiderict product we must have $|N \rtimes H| = |N| \times |H|$.