Answer
This set of vectors does have a vector space structure.
Work Step by Step
Let $u, v$ be in $\mathbb{R}^2$. So, $u=(u_1, u_2)$ and $v=(v_1, v_2)$ for some $u_1, u_2, v_1, v_2 \in \mathbb{R}$. $u+v = (u_1+v_1+1, u_2+v_2+1)$, so $u+v$ is in $\mathbb{R}^2$. Therefore, the first axiom holds.
Let $u \in \mathbb{R^2}$ and let $k \in \mathbb{R}$. $ku$ can quickly be checked to see that $ku$ is still a vector of length $2$, so $ku \in \mathbb{R}^2$.
Since both axioms hold, the space is a vector
