**Edited Answer (with minor corrections from comments below)**

I think this answer is morally the same as @fedja's above who beat me to it, but maybe slightly more explicit. Set $f(x)=x^5\sin(\pi/x)/(1+|x|^5)$, so that it looks like $x^5\sin(\pi/x)$ near the origin and define $h(x,y)=(x,y+\epsilon f(x))$. This is a global $C^2$ diffeomorphism and satisfies $\|D_xh(v)\|\ge c_\epsilon\|v\|$ for all $x$ and $v$ for a uniform $c_\epsilon$. Choose $\epsilon$ such that $c_\epsilon>1/\sqrt 2$. Now set $T(x,y)=(2x,2048y)$ and consider $\tilde T=h^{-1}\circ T\circ h$ (so that $\tilde T$ is globally expanding). Iterates of the $x$-axis under $\tilde T$ are conjugate by $h$ to iterates of the graph of $f$ under $T$, so it suffices to show that iterates of $\Gamma_f$, the graph of $f$, under $T$ become dense in the unit ball. 

Let $2^n>2/\epsilon$ and consider the image of $\Gamma_f$ under $T^n$. Let $k$ be in the range $n$ to $2n-1$. $\Gamma_f$ crosses the $x$ axis $2^k$ times in $(2^{-(k+1)},2^{-k}]$ with crossings roughly $2^{-2k}$-dense. The height of the graph is approximately $\pm 2^{-5k}\epsilon$ between each pair of zeros. Iterating $n$ times, the image of the graph reaches height $2^{11n}\times \pm 2^{-5k}\epsilon$ between each pair of zeros, so that the $n$th iterate of the graph crosses $2^k$ times between roughly $\pm 2^{11n-5k}\epsilon$ (since $2^{11n-5k}>2^n$, the $n$th iterate of the part of $\Gamma_f$ with $x$ values between $2^{-(k+1)}$ and $2^{-k}$ covers $[-1,1]$) for $x$ values that are $2^{n-2k}$-dense (so at least $2^{-n}$ dense) in each range $[2^{n-(k+1)},2^{n-k}]$ with $k=n,\ldots,2n-1$. That is the image of $\Gamma_f$ under $T^n$ contains "branches" that cover $[-1,1]$ and that are $2^{-n}$ dense in the range $\pm[2^{-n},1]$. In particular, the $T$ iterates of $\Gamma_f$ become dense in the unit ball, so the $\tilde T$ iterates of the $x$-axis become dense in the unit ball.