This is false. Let $G$ be the open subset $\textbf{GL}_2$ in the affine space $\textbf{Mat}_{2\times 2} = \mathbb{A}^4$ with affine coordinates $x_{1,1}$, $x_{2,1}$, $x_{1,2}$ and $x_{2,2}$. Let $V$ be the one-dimensional subvariety, $$ V = \text{Zero}(x_{2,1},x_{2,2}-1,x_{1,1}-x_{1,2}) = \left\{\left[ \begin{array}{rr} s & s \\ 0 & 1 \end{array} \right] : s\in \mathbb{A}^1 \right\}. $$ Similarly, let $W$ be the following one-dimensional subvariety, $$ V = \text{Zero}(x_{2,1},x_{1,1}-1,x_{2,2}-x_{1,2}) = \left\{\left[ \begin{array}{rr} 1 & t \\ 0 & t \end{array} \right] : t\in \mathbb{A}^1 \right\}. $$ Each of $V$ and $W$ is a one-dimensional, affine linear variety. In particular, each has degree $1$. Yet the product set $\overline{V\cdot W}$ is the following two-dimensional variety, $$ W = \text{Zero}(x_{2,1},2x_{1,1}x_{2,2}-x_{1,2}) = \left\{\left[ \begin{array}{rr} s & 2st \\ 0 & t \end{array} \right] : (s,t)\in \mathbb{A}^2 \right\}. $$ This has degree $2$.