Using integration by substitution you easily deduce $$\int_{-\infty}^\infty \exp(-ax^2) dx = \sqrt{\frac{\pi}{a}}$$ for $a>0$ from the case where $a=1$.
Question 1: Is there a similar way to prove the same formular for not only real $a>0$ but for $a \in \mathbb C$ with $\Re(a)>0$? Here I don't want to make use of the identity theorem.
Question 2: If you want to prove the equation using the identity theorem you need to argue that $$ a \mapsto \int_{-\infty}^\infty \exp(-ax^2) dx$$ is holomorphic. How do you argue that?
Side note: WolframAlpha thinks that the equality even holds for $a=i$. See here.
Assume that $a\in\mathbb{C},\,\text{Re}(a)>0.$. Then note that, integrating by parts, we have $$\int_{-\infty}^{+\infty}\exp\left(-ax^{2}\right)dx=2\int_{-\infty}^{+\infty}ax^{2}\exp\left(-ax^{2}\right)dx.$$Now define $$I\left(a\right):=\int_{-\infty}^{+\infty}\exp\left(-ax^{2}\right)dx.$$Then $$I^{\prime}\left(a\right)=\int_{-\infty}^{+\infty}\frac{\partial}{\partial a}\exp\left(-ax^{2}\right)dx=-\int_{-\infty}^{+\infty}x^{2}\exp\left(-ax^{2}\right)dx=-\frac{I\left(a\right)}{2a}$$ hence, solving the ODE, we get$$I\left(a\right)=\frac{c_{1}}{\sqrt{a}}$$ and so now it is enough to observe that $I\left(1\right)=c_{1}=\sqrt{\pi}$.