Is $\mathbb{R}^2$ with scalar multiplication only applying to the first element a vector space
I agree with Gary's comment. Excellent way to show that under this product $\mathbb R^2$ is not a vector space.
Here is an alternative way, I am assuming that the sum of two vectors in $\mathbb R^2$ is defined as usual sense i.e. $(x,y)+(u,v)=(x+u,y+v)$, and the scalar product is defined as $c(x,y)=(cx,y)$. Then the following contradiction occurs, $$(1,1)+(1,1)=(2,2)\ne (2,1)=2\times(1,1).$$