Consider the map $f: \Bbb R^2 \times \Bbb R^2 \to \Bbb R^4$ given by $(ae_1 + be_2, ce_1+de_2) \mapsto (ac, ad, bc, bd)$. This can be checked to be bilinear, which by the universal property of the tensor product reduces the given problem to showing that $f(e_1, e_1) + f(e_2, e_2)$ is not of the form $f(v,w)$ for any $v,w$.
(Here the intution is that the induced map $\Bbb R^2 \otimes \Bbb R^2 \to \Bbb R^4$ is actually the isomorphism which identifies the standard basis of $\Bbb R^4$ with the usual basis of $\Bbb R^2 \otimes \Bbb R^2 $)
Now note that $f(e_1, e_1) = (1,0,0,0)$ and $f(e_2, e_2) = (0,0,0,1)$. If we set $v = ae_1 + be_2, w = ce_1+de_2$, then $f(v,w) = (ac, ad, bc, bd) = (1,0,0,1)$ has no solutions, since if $ad=0$ then either $a=0$ or $d=0$, which contradict $ac = 1, bd=1$, respectively.