由题意得\[ \begin{cases}a=2,\\ {\dfrac{c}{a}}={\dfrac{{\sqrt{2}}}{2}},\\a^2=b^2+c^2,\end{cases} \]解得\[ b={\sqrt{2}}. \]所以，椭圆 $ C $ 的方程为\[ {\dfrac{x^2}{4}}+{\dfrac{y^2}{2}}=1. \]

由\[ \begin{cases} y=k\left(x-1\right),\\ {\dfrac{x^2}{4}}+{\dfrac{y^2}{2}}=1, \end{cases} \]得\[ \left(1+2k^2\right)x^2-4k^2x+2k^2-4=0. \]设点 $M $，$ N$ 的坐标分别为 $ \left(x_1,y_1\right)$，$\left(x_2,y_2\right) $，则\[ y_1=k\left(x_1-1\right),y_2=k\left(x_2-1\right),\\ x_1+x_2={\dfrac{4k^2}{1+2k^2}},x_1x_2={\dfrac{2k^2-4}{1+2k^2}}.\]所以\[ \begin{split}|MN| &={\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}}\\&={\sqrt{\left(1+k^2\right)\left[\left(x_1+x_2\right)^2-4x_1x_2\right]}}\\&={\dfrac{2{\sqrt{\left(1+k^2\right)\left(4+6k^2\right)}}}{1+2k^2}}.\end{split} \]又因为点 $ A\left(2,0\right) $ 到直线 $ y=k\left(x-1\right) $ 的距离\[ d={\dfrac{|k|}{{\sqrt{1+k^2}}}}, \]所以 $ \triangle AMN $ 的面积为\[ S={\dfrac{1}{2}}|MN|\cdot d={\dfrac{|k|{\sqrt{4+6k^2}}}{1+2k^2}}. \]由 $ {\dfrac{|k|{\sqrt{4+6k^2}}}{1+2k^2}}={\dfrac{{\sqrt{10}}}{3}} $，解得\[ k=\pm 1. \]