If two positively oriented bases define the same lattice, then there is a matrix in $\operatorname{SL}(2,\mathbb{Z})$ acting on this basis Remark: This is exactly Exercise 3.7.2 (3) from Lozano-Robledo's book Elliptic Curves, Modular Forms and their L-functions.
Definition: A lattice in the complex plane is a set of the form $$\langle w_1, w_2 \rangle := \{ m w_1 + n w_2 | m,n \in \mathbb{Z} \}$$ where $w_1, w_2 \in\mathbb{C}$ are linearly independent when interpreting them as vectors in $\mathbb{R}^2$.
Problem: Let $L=\langle w_1,w_2\rangle$, $L'=\langle w_1',w_2'\rangle$ be lattices with positively oriented bases, i.e. the imaginary parts of $\tau := \frac{w_1}{w_2}$, $\tau' := \frac{w_1'}{w_2'}$ are positive. I want to show that if $L = L'$ then there is a matrix $M \in \operatorname{SL}(2,\mathbb{Z})$ such that $M \begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = \begin{pmatrix} w_1' \\ w_2' \end{pmatrix}$.
My progress: Because $L = L'$ there exist integers $a,b,c,d \in \mathbb{Z}$ with $w_1' = a w_1 + b w_2$ and $w_2' = c w_1 + d w_2$. That means we have 
$N \begin{pmatrix} w_1 \\ w_2 \end{pmatrix} := \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = \begin{pmatrix} w_1' \\ w_2' \end{pmatrix}$.
Now we need to find appropriate integers such that $ad - bc = 1$, in addition to the condition before, so this matrix  $N$ is really in $\operatorname{SL}(2,\mathbb{Z})$.
We notice that $N \tau = \frac{a \tau + b}{c \tau + d} = \tau'$.
Furthermore, one can show $\operatorname{Im}(\tau') = \frac{(ad-bc)\operatorname{Im}(\tau')}{|c \tau + d|^2} $. Because our basis is positively oriented, we can deduce that $ad-bc$ is a positive integer.
I thought that one could use the greatest common divisor to transform this equation to $ad-bc=1$. But I do not know how to elaborate this and moreover, I do not know if the conditions before are still satisfied.
Could someone help me please? Thank you!
Edit: I also thought about what would happen if c and d are coprime. Then there are integers $x, y: xc + yd =1$, then ideally $a = y, b=-x$. But I have no idea how I could use this observation. 
Edit 2: I also noticed that I have not used the fact that both lattices are equal completely. In particular, I have just used that $w_1', w_2'$ are generated by $w_1, w_2$. I think it is a good idea to utilize the full potential of the conditions, so let us assume that there are integers $e,f,g,h$ with $w_1 = e w_1' + f w_2'$ and $w_2 = g w_1' + h w_2'$ as well. Then we can show
$$ \operatorname{Im}(\tau') = \frac{(ad-bc) \operatorname{Im}(\tau)}{|c \tau + d|^2} = \frac{(ad-bc)(eh-fg) \operatorname{Im}(\tau')}{|(c \tau +d)(g \tau' + h)|^2 } $$,
so $\frac{(ad-bc)(eh-fg)}{|(c \tau +d)(g \tau' + h)|^2 } = 1$. Maybe this is something else one could use for the solution.

Hint: $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}e&f\\g&h\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}$$
and both factors are integer matrices. Their determinants are therefore integers; you have found them to be positive; and their product is $1$.
