This can be proved using the modular diagram in the upper half plane, and the dual tree $T$ of the modular diagram (see below for a discussion of the picture in the link). The modular diagram has vertices at the rational numbers together with $\infty = \frac{1}{0}$. The hyperbolic line $\overline{\frac{p}{r}, \frac{q}{s}}$ is an edge if and only if $ps-qr=\pm 1$ (all rational numbers are in lowest form).
The dual tree $T$ is bipartite, its $0$-cells of valence 2 vertices being where it crosses the edges of the modular diagram, and its $0$-cells of valence 3 vertices being the barycenters of the ideal triangles of the modular diagram. The $0$-cell of valence 2 on the side $\overline{\frac{1}{0},\frac{0}{1}}$ is
$$v_0 = 0 + 1i
$$
and $v_0$ is connected by a 1-cell, denoted $E_+$, to a 0-cell of valence 3 vertex at
$$v_1 = \frac{1}{2} + \frac{\sqrt{3}}{2} i
$$
Let $E_-$ denote the reflection of $E_+$ across the $y$-axis, joining $v_0$ to the point $-\frac{1}{2} + \frac{\sqrt{3}}{2} i$.
Consider paths in $T$ with initial edge $E_+$ that terminate on a $0$-cell of valence 2 (that $0$-cell having positive real part). These paths can be identified with words in the letters $L$ and $R$. For example, the path $RLRL$ starts at $v_0$, goes along $E_+$, then turns Right at the valence 3 vertex $v_1$, then continues past the next valence 2 vertex located at $\frac{1}{2} + \frac{1}{2} i$ to turn Left at the next valence 3 vertex, then Right, then Left, and then stops at the next valence 2 vertex.
What to show is the following list of properties of the fractional linear action of $GL(2,\mathbb{Z})$ restricted to $T$:
$\pmatrix{2 & 3 \\ 3 & 5}$ has an axis with fundamental domain $RLRL$
$\pmatrix{5 & 3 \\ 3 & 2}$ has an axis with fundamental domain $LRLR$
$\pmatrix{1 & n \\ 0 & 1}$ has an axis with fundmantal domain $\underbrace{LLLL…L}_{\text{$n$ times}}$
By constructing the entire axes of these elements (just continue forward and backward in $T$ with further copies of the fundamental domain), one sees that the intersection of the axis of $\pmatrix{2 & 3 \\ 3 & 5}$ and the axis of $\pmatrix{5 & 3 \\ 3 & 2}$ is therefore equal to $E_- \cup E_+$.
From this, it follows easily that the entire minimal subtree of the free group $\langle A,B\rangle$ intersects the axis of $\pmatrix{1 & n \\ 0 & 1}$ solely in a finite segment, namely the union of the segment $E_- \cup E_+$ connecting $-\frac{1}{2} + \frac{\sqrt{3}}{2} i$ to $v_1 = \frac{1}{2} + \frac{\sqrt{3}}{2} i$ together with the translate of that segment one unit to the right, connecting $v_1$ to $\frac{3}{2} + \frac{\sqrt{3}}{2} i$.
From this it follows that $\pmatrix{1 & n \\ 0 & 1} \not\in \langle A,B\rangle$ for all $n \ne 0$.
What remains is to show 1 and 2 above (3 being obvious).
The general statement is as follows. Denote
$$L = \pmatrix{1 & 1 \\ 0 & 1} \quad R = \pmatrix{1 & 0 \\ 1 & 1}
$$
Every matrix $\pmatrix{p & q \\ r & s} \in SL(2,\mathbb{Z})$ with $p,q,r,s > 0$ can be factored uniquely as a product of $L$ and $R$; simply use column reduction. The word you get from that factorization, when interpreted as a path in $T$ starting with $E_0$, is a fundamental domain for the axis of $\pmatrix{p & q \\ r & s}$.
So what's left is a column reduction calculation, to verify that
$$\pmatrix{2 & 3 \\ 3 & 5} = RLRL, \qquad \pmatrix{5 & 3 \\ 3 & 2} = LRLR
$$
Added bonus: Apropos of some of the comments, one can show, by examining the minimal subtree of $\langle A,B \rangle$ constructed above, that every ray in the minimal subtree oscillates infinitely often between $L$ and $R$. Therefore, this group has no cusps, because each ray in $T$ that limits on a cusp is eventually $RRRRRRR…$ or $LLLLLLL….$.
Added: I finally found a good picture on the web, linked to in the first line of the answer. The hyperbolic lines of the modular diagram are in blue, the dual tree $T$ is in red, and one can label the cusps along the bottom of the picture as $-1,0,1,2,3$.
