You have some kind of error in your formula for $d(e^x,e^y)$ because $d(e^x,e^x)$ is not identically zero. You need to replace $x$ by $-x$ in the right hand side of (3) to even have a chance of it being correct. However, I'm pretty sure that still won't produce a correct formula.

Yes, but you do have to at least assume that the $14$-piece of $\mathrm{d}({*}\Phi)$ vanishes in order to draw your conclusion.

Are you assuming that $\Phi$ is closed and co-closed? If so, then, yes, what you have written is an identity. If not, then it depends on the torsion of the $\mathrm{G}_2$-structure.

@PeterLeFanuLumsdaine: I forgot to say what this has to do with curvature: As one knows, the Riemann curvature tensor of a metric $g$ is a section of $S^2(\Lambda^2(T^*M))$. Since it is second order in $g$, which is a section of $S^2(T^*M)$, the curvature must be computable from the second derivatives of $g$, i.e., from something like $S^2(T^*M)\otimes S^2(T^*M)$. By the time one cuts away the coordinate (or connection) dependence of this section, one is left with some quotient of $S^2(S^2(T^*M))$, and the only part that can match with the Riemann curvature tensor is the $K(T^*M)$ component.

@PeterLeFanuLumsdaine: Oh, and I should have pointed out that, since $S^2(V^*)$ is canonically isomorphic to $\bigl(S^2(V)\bigr)^*$, we also have $S^2\bigl(S^2(V^*)^*\bigr)$ is canonically isomorphic to $S^2(S^2(V))$.

@PeterLeFanuLumsdaine: There's not that much to say. The usual multiplicity formulae in $\mathrm{GL}(V)$-representation theory show that $S^2(\Lambda^2(V))\simeq \Lambda^4(V)\oplus K(V)$ while $S^2(S^2(V))\simeq S^4(V)\oplus K(V)$, where $K(V)$ is irreducible. The center map of the exact sequence on decomposable elements is $$(v_1\wedge v_2)\circ(v_3\wedge v_4)\longmapsto (v_1\circ v_3)\circ(v_2\circ v_4)- (v_1\circ v_4)\circ(v_2\circ v_3).$$ It's clearly nonzero, so it has to match the two copies of $K(V)$, making $\Lambda^4(V)$ be the kernel and $S^4(V)$ be the cokernel.

Also, for another way that this $\mathrm{GL}(V)$ modules shows up, see my answer at mathoverflow.net/q/100372.

@AntonPetrunin: Yes, of course that's true. It's interesting that $K(V)$ is irreducible as a $\mathrm{GL}(V)$-module, whereas it's reducible as an $\mathrm{SO}(V)$-module (which is the way we usually think of it).