Tyler's comment to my earlier answer seems to give a solution; he suggests comparing the space $T=(S^3\vee S^3)\cup_{[x,[x,y]]} e^8$ with a wedge $S^3\vee S^3\vee S^8$. It's probably easier to think about homology with the Pontryagin product. Homology of loops on on the wedge will be a tensor algebra on classes in 2,2,7 (since its it's loops of a suspension). The homology of loops on Tyler-space $T$ should differ in dimension 6: the homology class [x,[x,y]] will be 0 (where x,y are now the homology generators in dimension 2), "killed" by the new attaching map. So H_6 (and thus H^6) of the two spaces have different rank.
To make this explicit, we have $S^7 -f-> \xrightarrow{f} X --> T\rightarrow T$, where X is the wedge of two 3-spheres. The restriction of Omega $\Omega f: \Omega S^7 --> \to \Omega X X$ is a map $S^6 --> \to \Omega X X$ adjoint to f, and on homology this hits the homology class corresponding to the [x,[x,y]]. The result follows because Omega $\Omega S^7 --> \to \Omega X --> \to \Omega T T$ is null homotopic. (I'm basically using the Hilton-Milnor theorem to understand Omega X.$\Omega X$.)
