I am trying to prove the previous lemma and I have thought to do the following: To go from right to left, I do not know how to proceed, but to go from left to right, it occurs to me to do the following.
Suppose that there is such a matrix, this means that $w_1'=aw_1+bw_2$ and $w_2'=cw_1+dw_2$, with which $\mathbb{Z}w_1'\oplus\mathbb{Z}w_2'=\mathbb{Z}(aw_1+bw_2)\oplus\mathbb{Z}(cw_1+dw_2)=\mathbb{Z}w_1\oplus\mathbb{Z}w_2$. Is this reasoning right? How do I prove the other implication? Thank you.
For $\omega_1'$, $\omega_2'$ to be a basis of $\Lambda$ it's necessary that they are $\Bbb R$-linearly independent elements of $\Lambda$, that is $$\pmatrix{\omega_1'\\\omega_2'}=A\pmatrix{\omega_1\\\omega_2}$$ where $A$ is a nonsingular integer matrix.
In this case $\omega_1'$, $\omega_2'$ do form a basis of some lattice $\Lambda'$. For this to be the same as $\Lambda$ we need $\omega_1$, $\omega_2\in\Lambda'$, so that $$\pmatrix{\omega_1\\\omega_2}=B\pmatrix{\omega_1'\\\omega_2'}$$ for some integer matrix $B$. Then $BA=I$ which implies, $\det(B)\det(A)=1$ and so $\det A=\pm1$. But if $\det A=-1$ one shows that the sign of the imaginary part of $\omega_1'/\omega_2'$ is opposite that of $\omega_1/\omega_2$. Therefore $A\in\text{SL}_2(\Bbb Z)$.