Perhaps the following explicit calculations clarify Ekedahl's argument: consider, for example, the case where the group is cyclic of order 3. Let $w$ be a generator of the group. The characteristic polynomial of w is then necessarily $X^2 + X + 1$. Thus the action of $w$ on $M = \mathbf{Z}\oplus\mathbf{Z}$ makes $M$ into a $\mathbf{Z}[X]/(X^2 + X + 1)$-module. This module is torsion free (since it is torsion-free as a $\mathbf{Z}$-module). Consequently, it is locally free (necessarily of rank $1$) and hence free since $\mathbf{Z}[X]/(X^2+X+1)$ is a PID.

The freeness means that one can find a module generator $m\in M$. Since $(-X,1)$ is a $\mathbf{Z}$-module basis for $\mathbf{Z}[X]/(X^2+X+1)$, one finds that that $(-wm,m)$ is a $\mathbf{Z}$-module basis for $M$.

If the basis $(-wm,m)$ has the same orientation as the standard basis for $M$, then the change-of-basis matrix to $(-wm,m)$ is in ${\rm SL}_2(\mathbf{Z})$. Since the matrix of $w$ with respect to $(-wm,m)$ is the standard matrix of the original problem statement, this construction finishes the argument in this case.

If, on the other hand, the basis $(-wm,m)$ has the opposite orientation from the standard basis for $M$, then $(-w^2 m, m)$ has the same orientation as the standard basis of $M$ (algebra omitted). The change-of-basis matrix to $(-w^2 m, m)$ is thus in ${\rm SL}_2(\mathbf{Z})$. The matrix of $w^2$ with respect to $(-w^2 m, m)$ is the standard matrix of the original problem statement, which finishes the argument in this case.

Let me state this last part of the calculation abstractly: since the question involves ${\rm SL}_2(\mathbf{Z})$ conjugacy, one should consider locally free $\mathbf{Z}[X]/(X^2+X+1)$-modules of rank $1$ equipped with an orientation. We use the orientation $(-X,1)$ on $\mathbf{Z}[X]/(X^2+X+1)$. Given such a module $N$, we can obtain a new *conjugate module* $N'$ by changing the $\mathbf{Z}[X]/(X^2+X+1)$ action by the (orientation-reversing) automorphism $X\mapsto X^2$. For any oriented $N$, either $N$ or $N'$ is module-isomorphic to $\mathbf{Z}[X]/(X^2+X+1)$ (with orientation). Corresponding to these two cases, one can conjugate (in ${\rm SL}_2(\mathbf{Z})$) either $w$ or $w^2$ to the desired standard form.

Note that it is straightforward to find a module generator $m$ (and hence to find a standardizing basis for $M$): the function $$ q(m):m\mapsto \det(-wm \; m) $$ is a quadratic form on $\mathbf{Z}\oplus\mathbf{Z}$. It is either positive definite or negative definite. (One can see that easily in the abstract using a module generator for $M$.) An $m\in M$ is a module generator precisely when $q(m) = \pm 1$. We know that such an $m$ exists, and since $q$ is definite, it is easy to search for one.