A conceptual explanation (since the other answers have given counterexamples) for why this is untrue is as follows: A sheaf consists of local data and global data specifying how those local data fit together. Even if all of the local data of two sheaves are isomorphic, there is no reason to believe that those isomorphisms can be fit together in a compatible way. This is why we require that the isomorphisms on stalks arise from a map that is already a morphism of sheaves, since this exactly says that the data fit together in the proper way.
We even encounter the same problems when we work with presheaves, for instance. Since presheaves are functors, a morphism of presheaves must be a natural transformation of functors. However, simply having isomorphisms "pointwise", as it were, is not enough. The isomorphisms must also commute with the restriction maps.
