FALSE: the same delta works after every finite change of generating set

The generating set $\{a,b\}$ gives a tree Cayley graph, so $F_2$ is $0$-hyperbolic for that choice.

Let $\delta \ge 0$, and choose an even $n=2k$ with $k>\delta$. The paths labelled $a^n$ from $1$ to $a^n$ and $b^n$ from $a^n$ to $t$ have length $n$. They are geodesic: for every integer $r$, [A1] gives respective lower bounds $\mathrm{\mid }n-nr\mathrm{\mid }+\mathrm{\mid }nr\mathrm{\mid }+\mathrm{\mid }r\mathrm{\mid }\ge n$ and $\mathrm{\mid }nr\mathrm{\mid }+\mathrm{\mid }n-nr\mathrm{\mid }+\mathrm{\mid }r\mathrm{\mid }\ge n$. The edge labelled $t$ joins $1$ to $t$, so these paths form a geodesic triangle.

Its midpoint vertex $a^k$ is distance $k$ from $1$. For any word from $a^k$ to $t$, [A1] gives $\mathrm{\ell }\ge \mathrm{\mid }k-nr\mathrm{\mid }+\mathrm{\mid }n-nr\mathrm{\mid }+\mathrm{\mid }r\mathrm{\mid }\ge k+1$ for every integer $r$: the first term is at least $k$, and either $r\mathrm{\ne }0$ or the second term is positive. Thus $a^k$ is at distance at least $k$ from the one-edge side $[1,t]$, including its interior. Every vertex $a^n{b}^j$ on the other long side, $0\le j\le n$, is also at distance at least $k$ from $a^k$, since the first term of the lower bound $\mathrm{\mid }k-nr\mathrm{\mid }+\mathrm{\mid }j-nr\mathrm{\mid }+\mathrm{\mid }r\mathrm{\mid }$ is at least $k$ for every integer $r$. The same bound holds for points inside its edges. Therefore this geodesic triangle is not $\delta$-slim.

Since $\delta$ was arbitrary, no single constant works for all finite generating sets of $F_2$. Therefore the claim is false.