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Consider $S^1 \hookrightarrow M_f$ where $f:S^1 \rightarrow S^1$ is squaring and $M_f$ denotes the mapping cylinder. Then the inclusion induces a map $\pi_1(S^1) \rightarrow \pi_1(M_f)$ which has image $2 \mathbb{Z}$. Attaching a 2 dimensional disk to a graph has the effect of quotienting out by the normal subgroup generated by the map of the boundary. For the inclusion to remain an isomorphism, this means that any 2-disk we attach must be through a null homotopic map, since we are after a quotient of $\mathbb{Z}$ isomorphic to $\mathbb{Z}$. Obviously attaching a disk like this does nothing to solve the problem, so this is a counter example.