I suggest using the following characterization of ergodicity:
$\varphi$ is ergodic if and only if every $f\in L^2(\mathbb{T}^2)$ such that $f\circ \varphi = f$ is constant function.
Now let's use this criterion to prove $\varphi$ is ergodic. Suppose $f\in L^2$ with $f\circ \varphi = f$. Decompose $f$ into its Fourier series $f = \sum_{m,n\in \mathbb{Z}} = \alpha_{(m\,n)}e^{2\pi imx}e^{2\pi iny},$ with coefficients $\alpha_{(m\,n)}\in \mathbb{C}$. Then, if $\varphi$ is given by the matrix $A = \left(\begin{matrix} a & b\\c & d\end{matrix}\right),$ it is easy to compute that $f\circ \varphi = \sum_{m,n}\alpha_{(m\,n)}e^{2\pi i(ma+nc)x}e^{2\pi i(mb+nd)y}.$ Since $f$ is invariant, the Fourier series for $f$ and $f\circ \varphi$ must agree, so $\alpha_{(m\,n)} = \alpha_{(ma+nc\,mb+nd)}$ for all $m,n\in \mathbb{Z}$. We can express this more simply as follows. If $v = (m\,\,n)\in \mathbb{Z}^2$, then $\alpha_v = \alpha_{vA}$. By iterating, $\alpha_v = \alpha_{vA^k}$ for each $k\in \mathbb{Z}$.
Suppose that $v\in \mathbb{Z}^2$. Either the sequence $vA^k$ of vectors with integer coordinates is periodic, or else $\|vA^k\|\to \infty$ as $k\to \infty$. Note that the first case cannot happen unless $v = 0$, since if $v = vA^k$ for some $k$, then $A^k$ would have $1$ as an eigenvalue, which contradicts the assumption of hyperbolicity. Thus either $v = 0$, or $\|vA^k\|\to \infty$. Suppose $v\neq 0$. Since $f\in L^2$, the coefficients $\alpha_{(m\,n)}\to 0$ as $\|(m\,\,n)\|\to \infty$, and thus $\alpha_v = \alpha_{vA^k}\to 0$, i.e., $\alpha_v = 0$. We have therefore shown that the only way $\alpha_v$ can be nonzero is if $v = 0$. The Fourier series for $f$ is then $f = \alpha_0$, so $f$ is a constant function.