Yes, they are equivalent. To show this first suppose that $(T, \otimes)$ is a tensor product of $V$ and $W$ with respect to definition 1, let $(e_i)_{i\in I}$ and $(f_j)_{j\in J}$ be bases of $V$ and $W$, respectively. Define a bilinear map $B\colon V \times W \to K^{(I \times J)}$ by $B(e_i, f_j) = \delta_{(i,j)}$, where $\delta_{(i,j)}$ denotes the element of $K^{(I\times J)}$ which maps $(i,j)$ to 1 and all other pairs to 0. Then there is a unique linear $F\colon T \to K^{(I\times J)}$ with $F \circ \otimes = B$. We have $F(e_i \otimes f_j) = \delta_{(i,j)}$ for all $(i,j) \in I \times J$, so $F$ is a surjection. Moreover, as the $\delta_{(i,j)}$ are linearly independent, so are the $e_i \otimes f_j$. Let $U$ be the subspace of $T$ spanned by $(e_i\otimes f_j)_{i,j}$, let $X$ be a complement of $U$, i. e. $T = U \oplus X$. Let $\phi\colon X \to K^{(I \times J)}$ be a linear map, and define $G\colon T \to K^{(I \times J)}$ by $G(u+x) = F(u) + \phi(x)$ for $u \in U$ and $x \in X$. Then for $v \in V$, $w \in W$ as we have $v \otimes w \in U$ it holds $G(v \otimes w) = F(v \otimes w) = B(v,w)$. By uniqueness of $F$ we must have $G = F$ for all $\phi$. This yields $X = 0$, so $U = T$ and as $(e_i \otimes f_j)$ is a basis of $U$, it is one of $T$.
Now suppose $(T,\otimes)$ fulfills definition 2. Let $B\colon V \times W \to X$ be bilinear. Let $(e_i)$ resp. $(f_j)$ be bases of $V$ resp. $W$. Then $(e_i \otimes f_j)$ is a basis of $T$, so $F(e_i \otimes f_j) := B(e_i, f_j)$ extends uniquely to a linear mapping $F\colon T \to X$, which obviously has $F \circ \otimes = B$. If $G\colon T \to X$ is linear with $G \circ \otimes = B$ then we have $G(e_i \otimes f_j) = B(e_i, f_j) = F(e_i \otimes f_j)$, but now, as two linear mappings agreeing on a basis are identical we have $F= G$, so $F$ is unique.