Is the wedge product of two harmonic forms harmonic?

It is easy to construct examples on Riemann surfaces of genus $>1$. Take any surface like this. Let $A$ and $B$ be two harmonic $1$-forms, that are not proportional. Then $A \wedge B$ is non-zero, but it vanishes at some point, since both $A$ and $B$ have zeros. At the same time a harmonic $2$-form on a Riemann surface is constant. Explicit examples of $1$-forms on Riemann surfaces can be obtained as real parts of holomorphic $1$-forms.

Note of course that the above example is complex, and Einstein just take the standard metric of curvature $-1$. If you want an example on a Ricci flat manifold you should take a $K3$ surface. It is complex and admits a Ricci flat metric. Now, its second cohomology has dimension $22$. Now it should be possible to find two anti-self-dual two-forms whose wedge product vanishes at one point on $K3$ but is not identically zero. This is because the dimension of the space of self dual forms is 19 which is big enought to get vanishing at one point

Generically, the wedge product of two harmonic forms will not be harmonic. It is harder to find examples than counter-examples. For example, on compact Lie groups with a bi-invariant metric or, more generally, on riemannian symmetric spaces, harmonic forms are invariant and invariance is preserved by the wedge product. In general, though, this is not the case.

According to Kotschick (see, e.g., this paper) manifolds admitting a metric with this property are called geometrically formal and their topology is strongly constrained. He has examples, already in dimension 4, of manifolds which are not geometrically formal.

Interestingly in (24) of hep-th/9603176 it is mistakenly claimed that the wedge product of harmonic forms is automatically harmonic. Because it is false we still do not know the predicted existence of those middle dimensional $L^2$ harmonic forms on these non-compact complete hyperkahler manifolds.