Can all "standard" properties of the tensor product be proven from the universal property? The tensor product is typically constructed in an existence proof by referring to a rather esoteric quotient space which "feels" hard to work with in general. The universal property of bilinear factorization feels much more natural and motivated - but is it sufficient to, for instance, prove that the tensor product is commutative or that a basis for $V\otimes W$ is $\{e_i\otimes f_i\}$ without referring to this construction?
 A: Yes, of course. After all universal objects are unique up to canonical isomoprphism so you should be able to do all (or at least most) abstractly.
For example $V\otimes W\cong W\otimes V$ follows immediatly from $V\times W\cong W\times V$ via $(v,w)\mapsto(w,v)$, which respects bilinearity.
That the $v\otimes w$ span $V\otimes W$ follows from the fact that their span would also have the universal property, hence the inclusion of their span into $V\otimes W$ is the identiy.
And that any linear identity among $\{v_i\otimes w_j\}_{i,j}$ produces a linear identity among the $v_i$ or among the $w_j$ should also follow quite fast because then everything factors over the quotient
A: Short answer: yes.
In what sense do you mean that the tensor product is commutative? That switching $V$ and $W$ in $V\otimes W$ gives a space isomorphic to $W\otimes V$? Then we can show this in the following way: $W\otimes V$ satisfies the same universal property as $V\otimes W$ so they must be isomorphic. Consider any bilinear map $f$ from $V\times W \to T$, then this naturally gives a map from $W\times V \to T$, which factors through $W\otimes V$.
Let me clarify with a commutative diagram.
$\require{AMScd} \displaystyle
\begin{CD}
V\times W @>f>> T\\
@VVg=(v,w)\mapsto(w,v)V        @|\\
W\times V @>f'(w,v)=f(v,w)>> T \\
@VV\varphi_{W\otimes V}V @|\\
W\otimes V @>f''>> T
\end{CD}$
where $\varphi_{W\otimes V}$ is the map such that $(W\otimes V, \phi_{W\otimes V})$ has the universal property with respect to $W\times V$ and $f''$ is the map such that $f'=f''\circ \varphi_{W\otimes V}$ by the universal property of $W\otimes V$.
Then $f=f''\circ(\varphi_{W\times V}\circ g)$.
Therefore $(W\otimes V,\varphi_{W\otimes V}\circ g)$ has the same universal property as $V\otimes W$, so the two spaces are isomorphic.
Similar things can be done to show the basis property as discussed in other answers.
A: Here is a nice way to see the fact about bases which generalizes to something useful as well. One way to state the universal property of the tensor product of vector spaces is that it is part of an adjunction
$$\text{Hom}(U \otimes V, W) \cong \text{Hom}(U, [V, W])$$
where $[V, W]$ denotes the vector space of linear maps $V \to W$. This says that $(-) \otimes V$ is a left adjoint (with right adjoint $[V, -]$), and hence it preserves colimits; in particular, the tensor product distributes over (possibly infinite) direct sums in both variables. 
Now, giving $U$ a basis is tantamount to writing it as a direct sum of copies of the $1$-dimensional vector space $k$ ($k$ the underlying field), and similarly for $V$, so just distribute over both of these direct sums and (once you show that $k \otimes k \cong k$, which also follows from the universal property) you've written $U \otimes V$ as a direct sum of copies of $k$ labeled by pairs of a basis vector of $U$ and a basis vector of $V$. 
This adjunction exists in great generality and so one can deduce many analogous statements about e.g. the direct sum decomposition of a tensor product of representations of a group, or vector bundles over a space, etc. 
