The slope of the line described in the question is $1$ and it passes through point $(2, 3)$, so its equation is $y = x + 1$. This line intersects with the $x$-axis at point $(-1, 0)$, and thus $(-1, 0)$ is a vertex of $\Gamma$. Hence, $a = 1$.

And since point $(2, 3)$ is on $\Gamma$, we know that $\frac{2^2}{1} - \frac{3^2}{b^2} = 1$, and then $b^2 = 3$.

Denote $c = \sqrt{a^2 + b^2}$. Then the eccentricity of $\Gamma$ is $\frac{c}{a} = \frac{\sqrt{a^2 + b^2}}{a} = 2$.