In Miyake's book, Modular Forms, Ch 2.6, thm 2.6.9, there is a statement which relate to Fourier expansion of the Eisenstein series.
Let $\Gamma$ be a Fuchsian group, $\chi$ a character of $\Gamma$ of finite order, and $k$ an integer. We suppose $\chi(-1)=(-1)^k$ if $-1 \in \Gamma$. Let $\Lambda$ be a subgroup of $\Gamma$, and $\phi$ be a meromorphic function of $\mathbb{H}$ which is a seed function for a the Poincare series. We write the Poincare series by
$$
F(z)=F_k(z;\phi,\chi,\Lambda,\Gamma)=\sum_{\gamma \in \Lambda \setminus \Gamma} \overline{\chi(\gamma)} (\phi|_k \gamma)(z).
$$
Let $x$ be a cusp and $\sigma \in SL_2(\mathbb{R})$ satisfies $\sigma(x)=\infty$, and $h$ be a cusp width of $x$. Take $\phi_m(z)=j(\sigma,z)^{-k}e^{2\pi i m \sigma z/h}$, $\Lambda=\Gamma_x$, and take $\chi$ which satisfies
$$
\chi(\gamma)j(\sigma \gamma \sigma^{-1},z)^k =1
$$
for $\gamma \in \Gamma_x$. If $m=0$, then $F(z)=F_k(z;\phi_0,\chi,\Gamma_x,\Gamma)$ is called the Eisenstein series.
Theorem. Suppose $k\geq 3$. If $m=0$, then $F(z)=F_k(z;\phi_0,\chi,\Gamma_0,\Gamma) \in M_k(\Gamma,\chi).$ It has the Fourier expansion at $x$ of the form
$$
(F|_k \sigma^{-1})(z)=1+\sum_{n=1}^{\infty} a_n e^{2\pi i n z/h},
$$
and vanishes at all cusps which are inequivalent to $x$.
But I don't know why the Fourier coefficient $a_0$ is $1$.
Here is my attempt.
$$
a_0=\frac{1}{h}\int_{z_0}^{z_0+h} (F|_k \sigma^{-1})(z) dz
$$
$$
=\frac{1}{h}\int_{z_0}^{z_0+h}F_k(z;1,\chi^{\sigma},\sigma\Gamma_x \sigma^{-1} ,\sigma \Gamma \sigma^{-1}) dz,
$$
where $\chi^{\sigma}(\sigma \gamma \sigma^{-1})=\chi(\gamma)$, and the above integral is
$$
=\frac{1}{h}\int_{z_0}^{z_0+h}\sum_{\gamma \in \Gamma_x \setminus \Gamma } \chi(\gamma)^{-1}j(\sigma \gamma \sigma^{-1},z)^{-k} dz.
$$
The last integral must be 1, and if $\gamma \in \Gamma_x$, we know that $\chi(\gamma)j(\sigma \gamma \sigma^{-1},z)^k=1$. But I don't know about the other $\gamma$.
Best Answer
It seems that you started off well enough. Maybe for clarity, note that we may get rid of $\sigma$, assume that $x = \infty$ and what one really wants to show is that for all characters $\chi$ of finite order, which acts trivially on $\Gamma_{\infty}$ we have $$ I = \int_{z_0}^{z_0+h} \sum_{1 \ne \gamma \in \Gamma_{\infty} \backslash \Gamma} \chi(\gamma) j(\gamma, z)^{-k} dz = 0 $$
One could do that directly (see below), but in fact Miyake already does that in the course of proving Theorem 2.6.7. I would first like to draw your attention to condition (v), right before the theorem. Note that for $\phi = 1$, if $\sigma^{-1} = \left( \begin{array}[cc] aa & b \\ c & d \end{array} \right)$, then we get different bounds when $c = 0$ and when $c \ne 0$. Namely, for $c = 0$, we have $\varepsilon = 0$, but for $c \ne 0$, we have the much stronger $\varepsilon = k$.
Now, consider again the functions $\phi_{\alpha}$ in the proof of Theorem 2.6.7. In this case, $\alpha$ runs over a set of representatives for $\Gamma_{\infty} \backslash \Gamma / \Gamma_{\infty}$ .
Note that the lower left entry of a matrix (the one we call $c$) is the same for all elements of the double coset. Moreover, an element with $c = 0$ must be in $\Gamma_{\infty}$. Therefore, for any nontrivial $\alpha$, we have $c \ne 0$, and the stronger bound, implying that for any element $\alpha \beta$ in the double coset, this is the case, and so by the same proof as in that of Theorem 2.6.7, we see that $\phi_{\alpha}$ vanishes at $\infty$ for all $\alpha \ne \Gamma_{\infty}$.
Since $F(z) = \sum_{\alpha} \phi_{\alpha} (z)$ (see 2.6.6), we see that at $\infty$ the value of $F$ coincides with that of $\phi_1 = 1$.
(*) If you would really like to evaluate the integral, here is one way to proceed: \begin{align*} I &= \int_{\Gamma_{\infty} \backslash \mathbb{R}} \sum_{1 \ne \alpha \in \Gamma_{\infty} \backslash \Gamma / \Gamma_{\infty}} \sum_{\beta \in \Gamma_{\infty} \alpha \backslash \Gamma} \chi(\alpha \beta) j(\alpha \beta, z)^{-k} dz \\ &= \sum_{1 \ne \alpha \in \Gamma_{\infty} \backslash \Gamma / \Gamma_{\infty}} \chi(\alpha) \int_{\Gamma_{\infty} \backslash \mathbb{R}} \sum_{\beta \in (\alpha^{-1} \Gamma_{\infty} \alpha \cap \Gamma_{\infty}) \backslash \Gamma_{\infty}} j(\alpha, \beta z)^{-k} dz \\ &= \sum_{1 \ne \alpha \in \Gamma_{\infty} \backslash \Gamma / \Gamma_{\infty}} \chi(\alpha) \int_{(\alpha^{-1} \Gamma_{\infty} \alpha \cap \Gamma_{\infty}) \backslash \mathbb{R}} j(\alpha, z)^{-k} dz = 0. \end{align*}
Here we have used that $\chi(\beta) = 1$ for $\beta \in \Gamma_{\infty}$, that $j(\alpha \beta, z) = j(\alpha, \beta z) j(\beta, z)$, that $j(\beta, z) = 1$ for $\beta \in \Gamma_{\infty}$, that the integral $\int j(\alpha,z)^{-k} dz$ converges for $\alpha \notin \Gamma_{\infty}$, and that the sum $\sum \chi(\alpha)$ vanishes. This only works when $\chi$ is non-trivial due to convergence issues, but it gives a rough idea of what one should do.