I am trying to show that the Atkin-Lehner operator $W_q$ (as for example defined in Atkin and Lehner, 'Hecke operators on $\Gamma_0(N)$) is independent of the choice of entries, i.e. the operator $W_q$ is welldefined on the cusp forms $S_2(\Gamma_0(N))$.
Let $N$ be a positive integer.Let $q^\alpha$ be the exact power of $q$ dividing $N$, and let $x,y,z,w$ be any integers such that $W_q= \begin{pmatrix} q^{\alpha}x & y \\ Nz & q^\alpha w \end{pmatrix}$ has determinant $q$. For $f\in S_2(\Gamma_0(N))$ we denote, \begin{align*} W_qf = f\left \vert W_q\right. \end{align*} We want to show that the definition of operator $W_q$ does not depend on the choice of $x,y,z,w$ since for the matrices $W_q$ and $W_q'$ that satisfy the conditions of the definition. Suppose we have such a $W_q$ and $W_q'$. Then, \begin{align*} W_q W_{q}' = \begin{pmatrix} q^\alpha x & y \\ Nz & q^\alpha w \end{pmatrix} \begin{pmatrix} q^\alpha x' & y' \\ Nz' & q^\alpha w \end{pmatrix} = \begin{pmatrix} q^{2\alpha}x x'+yNz'& q^\alpha x y'+ ywq^\alpha \\ Nzq^\alpha x' +q^\alpha w N z' & Nzy'+ q^{2\alpha} w^2 \end{pmatrix}. \end{align*}
We see that $W_qW'_q$ has determinant $q^{2\alpha}$ and all entries are divisible by $q^\alpha$. So we can write $W_qW_q'=q^\alpha M$ for a matrix $M$ with determinant 1 and lower left entry $Nzx'+Nwz'$, so $M \in \Gamma_0(N)$. We also see that $W_{q'}^{-1}=W''_{q}$ satisfies the conditions of the definitions. So we have $W_q=q^\alpha M W_q''$ so we can write, for $f \in S_2(\Gamma_0(N))$, if we write $M=\left ( \begin{matrix} a & b \\ c & d \end{matrix} \right )$, \begin{align} f \vert W_q(z) = f \vert q^\alpha M W_q''(z) = (q^\alpha c z+dq^\alpha)^{-2}f((q^\alpha M)(z))\vert W''_q=q^{-2\alpha}f\vert W''_q(z) \end{align} where the last equality comes from the fact that $(q^\alpha M) z=Mz$ and $(cz+d)^{-2}f(Mz)=f(z)$ (since $M\in \Gamma_0(N)$).
Somehow we want the $q^{-2\alpha}$ to disappear, but I do not see how.
You write "We also see that $W_q'' = W_{q'}^{-1}$ satisfies the conditions of the definitions". You haven't defined anything called $W_{q'}$, so I think you mean $(W_{q}')^{-1}$; but this does not have integer entries and its determinant is $q^{-1}$, not $q$.
Besides, you're trying to show that $f \mid W_q = f \mid W_q'$; you've computed the action on $f$ of some other thing $W_{q}''$, but why does that solve the problem? A better way to approach this is to write $W_q' = A W_q$ for some matrix $A$, and then try to show that $A \in \Gamma_0(N)$.
(In this theory you also need to be careful how you normalise the action on modular forms of matrices of determinant other than 1; you need to define $$ f \mid \begin{pmatrix}a & b \\ c & d \end{pmatrix} = (ad - bc) (cz +d)^{-2} f( (az + b) / (cz + d) ).$$ If you omit the det factor, you get that $W_q^2$ is multiplication by $q^{2a}$, whereas you probably want it to be 1.)