It is enough to show two things:
1) The listed element generate the space.
We know that $(e_i \wedge e_j) \otimes e_k$ with no restraints on the indices generate the space, and we can clearly restrict to $i < j$ by the antisymmetric behaviour of the wedge product. Now, if $i > k$, we use the identity:
$$((u∧v)\otimes w=(w∧v)⊗u+(u∧w)⊗v$$
for $u=e_i$, $v=e_j$ and $w=e_k$.
This gives
$$((e_i∧e_j)\otimes e_k=(e_k∧e_j)⊗e_i+(e_i∧e_k)⊗e_j= (e_k∧e_j)⊗e_i-(e_k∧e_i)⊗e_j,$$
which shows that the vector is actually a linear combination of two of our given vectors. Therefore, they are indeed a generating set.
2) The listed elements are independent.
Note that the relations in the ideal are multilinear in $u$,$v$,$w$, so it is enough to use the relations for the basis vectors to generate the full ideal.
Now, suppose that you have a linear combination of the given vectors that gives zero in the quotient space. This means that the linear combination lies in the ideal which means that the linear combination is a linear combination of relations in the ideal.
You can now look at the smallest index of an $e_i$ that occurs in this expression and check that its coefficient is 0.