Finding the generators of a subgroup of $\mathrm{SL}_2(\mathbb Z)$

I am trying to solve the following problem:

Let $T_{ij}(c)\in\mathrm{SL}_2(\mathbb Z)\ (i\neq j)$ be the elementary matrix which represents the elementary row operation of adding the $j$-th row multiplied by $c$ to the $i$-th row, $A:=T_{12}(2)$ and $B:=T_{21}(2)$. Let $G$ be a subgroup $\left\{ \begin{pmatrix}a&b\\c&d\end{pmatrix}\in\mathrm{SL}_2(\mathbb Z):a\equiv d\equiv 1 \pmod 4, b\equiv c\equiv 0 \pmod 2 \right\}$. Show that $G = \langle A, B\rangle$.

That $\langle A,B\rangle\subset G$ can be shown easily by induction. The problem is the converse inclusion. I tried to show it by using an analogy from the fact that $\mathrm{SL}_2$ is generated by $T_{ij}(c)$. I tried to show that if $C\in G$ is written as a product $C_1\dots C_n$, where each $C_k$ is of the form $T_{i_kj_k}(c_k)$, then each $c_k$ is even, from which what I want to show follows. However I cannot prove this (it may simply be wrong).

I would be grateful if you could provide a clue.

I want to go by infinite descent, and to show that we can always make a given non-identity element of the group smaller by multiplying it with a generator or its inverse from the left or right. The trick is then to define "smaller" in a way that this is always possible. This is not guaranteed to work in general, but it is a natural thing to try.

Let $g=\pmatrix{a&b\cr c&d\cr}$ be an element of $G$. Let us first assume that $bd\neq0$. While that holds, I measure the size of $g$ by the absolute value of the product of the diagonal elements $|ad|$. The task at hand is then to prove that one of the matrices $Ag$, $A^{-1}g$, $Bg$, $B^{-1}g$, $gA$, $gA^{-1}$, $gB$, $gB^{-1}$ is smaller than $g$ in the prescribed sense. In other words we want to make $g$ smaller by adding one of its rows or columns to the other multiplied by $\pm2$ (multiplication by an elementary matrix from the right amount an elementary column operation).

W.l.o.g. we can assume that $|a|\ge |d|$ (make the obvious changes to the row/column operations below, if that is not the case. Then it is impossible that both inequalities $|b|>|a|$ and $|c|>|a|$ hold, as then $|bc|>|ad|$ and the congruences would imply that $|\det g|\ge3.$ So we have either $|b|<|a|$ or $|c|<|a|$. In the former case we can decrease $|a|$ by adding the second column multiplied by one of $\pm2$ to the first. As $|d|$ does not change we have the desired conclusion. In the latter case we add a multiply of the second row multiplied by $\pm2$ to the first for similar effect.

The above operation make $g$ smaller in the prescribed sense unless one of the off-diagonal elements is zero. I leave that case to you with the hint that if $b=0$ or $c=0$, then you can easily show that $a=d=1$.

The claim follows. The idea is that if $A$ and $B$ did not generate all of $G$, there would be a "minimal" missing element.