As usual, the functional equation on the Dirichlet series side comes from a theta function on the modular side.

Using the poisson summation formula we show $$\vartheta_z(x) = \sum_{(c,d) \in \mathbb{Z}^2} \exp(-\pi x \frac{|cz+d|^2}{|\Im(z)|}) = x^{-1} \vartheta_z(1/x)$$

Then let $$E_z(s) = \sum_{\gamma \in SL_2(\mathbb{Z})} \Im(\gamma(z))^{s}=\sum_{(c,d) \in \mathbb{Z}^2, gcd(c,d)=1} (\frac{|cz+d|^{2}}{\Im(z)})^{-s}$$
(with $gcd(0,d) = d$). It comes from a Mellin transform
$$\Gamma(s)\pi^{-s}\zeta(2s)E_z(s) = \Gamma(s)\pi^{-s}\sum_{c,d \in \mathbb{Z}^2 \setminus(0,0)} (\frac{|cz+d|^2}{|\Im(z)|})^{-s}\\ = \int_0^\infty (\vartheta_z(x)-1) x^{s-1}dx = \int_1^\infty (\vartheta_z(x)-1) x^{s-1}dx+\int_1^\infty (\vartheta_z(1/x)-1) x^{-s-1}dx \\ = \frac{1}{s-1}+\frac{1}{-s }+\int_1^\infty (\vartheta_z(x)-1) (x^{s-1}+x^{-s})dx$$
Which proves the functional equation.

Let
$h(u) = e^{-\pi \|u\|^2}, u \in \mathbb{R}^2$ which is its own Fourier transform. For some self-adjoint matrix $B \in GL_2(\mathbb{R})$ let $g(u)= e^{-\pi u^T B u} = h(B^{1/2}u)$ then $\widehat{g}(u) = \frac{1}{|\det(B)|^{1/2}} h((B^{-1/2})^Tu)$. Then apply the Poisson summation formula to
$$\theta_B(x) = \sum_{n \in \mathbb{Z}^2} e^{-\pi x (u^T Bu)}= \sum_{n \in \mathbb{Z}^2} h((x^{1/2} I)B n)\\=\sum_{n \in \mathbb{Z}^2} \frac{x^{-1}}{|\det(B)|^{1/2}} h((x^{-1/2}I)B^{-T/2}n)= \frac{x^{-1}}{|\det(B)|^{1/2}}\theta_{B^{-1}}(1/x)$$

Finally $|cz+d|^2 = (c,d)B{\scriptstyle\begin{pmatrix} c \\ d \end{pmatrix}}$ where $B = {\scriptstyle\begin{pmatrix} |z|^2 & \Re(z) \\ \Re(z) & 1\end{pmatrix}}, \det(B) = \Im(z)^2, B^{-1} = \frac{1}{\Im(z)^2} {\scriptstyle\begin{pmatrix} 1 & -\Re(z) \\ -\Re(z) & |z|^2\end{pmatrix}}$ so that
$(c,d)B^{-1}{\scriptstyle\begin{pmatrix} c \\ d \end{pmatrix}} = \frac{|c-dz|^2}{\Im(z)^2}$ and $\theta_{B^{-1}}(x) =\theta_B(\frac{x}{\Im(z)^2})$ and $\vartheta_z(x) = \theta_B(\frac{x}{|\Im(z)|})$.

(it is very possible there are some typos)