= Solution
A continuous map $F$ of a <compact metric space> is \b[<uniquely ergodic>] if it has exactly one <invariant measure> that is a <Borel probability measure>. The uniqueness requirement ranges over all invariant <Borel probability measures>.
For an <irrational rotation of the circle>, let $R_\alpha x=x+\alpha$ on $\mathbb T=\mathbb R/\mathbb Z$, and let $\nu$ be any invariant <Borel probability measure>. For each integer $r$, set $\chi_r(x)=e^{2\pi irx}$. Invariance gives
$$
\int\chi_r\,d\nu=\int\chi_r\circ R_\alpha\,d\nu
=e^{2\pi ir\alpha}\int\chi_r\,d\nu.
$$
For $r\ne0$, irrationality forces $e^{2\pi ir\alpha}\ne1$, so the corresponding <Fourier coefficient> is zero. For $r=0$ it is one. These are exactly the <Fourier coefficients> of normalized <Lebesgue measure> $m$. By the <Stone-Weierstrass theorem>, <trigonometric polynomials> are uniformly dense in the continuous functions on the <circle group>; hence $\nu$ and $m$ integrate every continuous function equally and are the same <Borel probability measure>. Since $m$ is invariant, \b[the <irrational rotation of the circle> is <uniquely ergodic>, with unique measure $m$].
For the <irrational skew shift> on the two-dimensional <torus>, write
$$
T(x,y)=(x+\alpha,y+x)\pmod1,\qquad
e_{r,s}(x,y)=e^{2\pi i(rx+sy)}\quad(r,s\in\mathbb Z).
$$
The map is invertible, with $T^{-1}(x,y)=(x-\alpha,y-x+\alpha)$ modulo one, and it preserves $m_2$ as allowed in the question. We first prove the <ergodic transformation> property by the <invariant-function characterization of ergodicity>.
Let $F\in L^2(m_2)$ satisfy $F\circ T=F$. Its expansion in the <Fourier basis> is $F=\sum_{r,s}c_{r,s}e_{r,s}$ in $L^2$. Direct calculation of the <Koopman operator> gives
$$
e_{r,s}\circ T=e^{2\pi ir\alpha}e_{r+s,s}.
$$
Uniqueness of the <Fourier coefficients> therefore implies
$$
c_{r,s}=e^{2\pi i(r-s)\alpha}c_{r-s,s}.
$$
For $s\ne0$, the magnitudes of the <Fourier coefficients> along all distinct indices $(r+js,s)$, $j\in\mathbb Z$, are equal. By the <Bessel inequality> they are square summable, so every such coefficient must be zero. When $s=0$, the relation becomes $c_{r,0}=e^{2\pi ir\alpha}c_{r,0}$, which forces $c_{r,0}=0$ for $r\ne0$. Only $c_{0,0}$ remains. Thus every invariant $L^2$ function is constant, and
$$
\boxed{(\mathbb T^2,\mathcal B,m_2,T)\text{ is ergodic}.}
$$
To prove <unique ergodicity>, we will establish \b[uniform averages for every continuous function] directly. No theorem on <unique ergodicity> of <skew products> is needed. Induction on $n$, using the old first coordinate in the second coordinate of $T$, yields
$$
\boxed{T^n(x,y)=\left(x+n\alpha,\ y+nx+\frac{n(n-1)}2\alpha\right)\pmod1.}
$$
In particular, for a <Fourier basis> element,
$$
u_n=e_{r,s}(T^n(x,y))
=\exp\left(2\pi i\left[rx+sy+n(r\alpha+sx)+\frac{s\alpha}{2}n(n-1)\right]\right).
$$
If $s=0$ and $r\ne0$, this is a <geometric series> in $n$, and
$$
\left|\frac1N\sum_{n=0}^{N-1}u_n\right|
\leq\frac{2}{N|1-e^{2\pi ir\alpha}|}\longrightarrow0
$$
uniformly in $(x,y)$.
For $s\ne0$, we give the finite <Van der Corput inequality for finite scalar sequences> and its proof. For $|u_n|\leq1$, extend $u_n$ by zero outside $0\leq n<N$. Fix an integer $1\leq H\leq N$ and set $C_h=\sum_{n=0}^{N-h-1}u_{n+h}\overline{u_n}$. Every original summand appears $H$ times in the identity
$$
H\sum_{n=0}^{N-1}u_n=\sum_{j=0}^{N+H-2}\sum_{a=0}^{H-1}u_{j-a}.
$$
The <Cauchy-Schwarz inequality>, followed by expansion of the squared window sums, gives
$$
H^2\left|\sum_{n=0}^{N-1}u_n\right|^2
\leq(N+H-1)\left(HN+2\sum_{h=1}^{H-1}(H-h)|C_h|\right).
$$
Consequently the <Van der Corput inequality for finite scalar sequences> is
$$
\boxed{\left|\frac1N\sum_{n=0}^{N-1}u_n\right|^2
\leq\frac{N+H-1}{NH}\left(1+2\sum_{h=1}^{H-1}\left(1-\frac hH\right)\frac{|C_h|}{N}\right).}
$$
The finite prefactor is important; the order of limits will be $N\to\infty$ with $H$ fixed, followed by $H\to\infty$.
For the <irrational skew shift> character sequence above, differencing cancels the quadratic term:
$$
u_{n+h}\overline{u_n}
=\exp\left(2\pi i\left[s\alpha hn+h(r\alpha+sx)+\frac{s\alpha}2h(h-1)\right]\right).
$$
For every fixed $h\geq1$, $s\alpha h$ is irrational, so another <geometric series> estimate gives
$$
\frac{|C_h|}{N}\leq\frac{2}{N|1-e^{2\pi is\alpha h}|}\longrightarrow0,
$$
uniformly in $(x,y)$. With fixed $H$, the <Van der Corput inequality for finite scalar sequences> therefore implies
$$
\limsup_{N\to\infty}\sup_{(x,y)\in\mathbb T^2}
\left|\frac1N\sum_{n=0}^{N-1}e_{r,s}(T^n(x,y))\right|^2\leq\frac1H.
$$
Letting $H\to\infty$ proves that the averages of every nonconstant <Fourier basis> element converge uniformly to zero. The constant character has average one. This establishes <uniform equidistribution of an irrational skew shift> on all <trigonometric polynomials>.
The <Stone-Weierstrass theorem> makes these <trigonometric polynomials> uniformly dense in $C(\mathbb T^2)$. If $p$ approximates a continuous $g$ with $\|g-p\|_\infty<\varepsilon$, then
$$
\sup_z\left|\frac1N\sum_{n=0}^{N-1}g(T^nz)-\int g\,dm_2\right|
\leq2\varepsilon+\sup_z\left|\frac1N\sum_{n=0}^{N-1}p(T^nz)-\int p\,dm_2\right|.
$$
Taking $N\to\infty$ and then $\varepsilon\downarrow0$ proves uniform convergence to $\int g\,dm_2$ for every continuous $g$.
Finally, if $\nu$ is any invariant <Borel probability measure> for the <irrational skew shift>, invariance and this uniform convergence give
$$
\int g\,d\nu
=\int\frac1N\sum_{n=0}^{N-1}g\circ T^n\,d\nu
\longrightarrow\int g\,dm_2\qquad(g\in C(\mathbb T^2)).
$$
Thus $\nu=m_2$, since continuous functions determine <Borel probability measures> on a <compact metric space>. We conclude
$$
\boxed{T\text{ is uniquely ergodic, with unique invariant probability }m_2.}
$$
The uniform-average proof also shows that every starting point has the same limiting continuous-function averages, a stronger conclusion than the almost-everywhere assertion provided by the <pointwise ergodic theorem>.
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