Claim — The graph $G'$ has at least as many triangles as $G$, and has strictly more if $G$ has any fatal vertices.
Proof. Obviously any triangle in $G$ persists in $G'$. Moreover, suppose $v$ is a fatal vertex of $G$. Then the neighbors of $G$ will form a clique in $G'$ which was not there already, so there are more triangles. $\square$

Thus we only need to consider graphs $G$ with no fatal vertices. Looking at the connected components, the only possibilities are cliques (including single vertices), cycles, and paths. So in what follows we restrict our attention to graphs $G$ only consisting of such components.

Remark (Warning). Beware: assuming $G$ is connected loses generality. For example, it could be that $G = G_1 \sqcup G_2$, where $G_1' \cong G_2$ and $G_2' \cong G_1$.

First, note that the following are stable under the operation:
* an isolated vertex,
* a cycle of odd length, or
* a clique with at least three vertices.

In particular, $G \cong G''$ holds for such graphs.

On the other hand, cycles of even length or paths of nonzero length will break into more connected components. For this reason, a graph $G$ with any of these components will not satisfy $G \cong G''$ because $G'$ will have strictly more connected components than $G$, and $G''$ will have at least as many as $G'$.

Therefore $G \cong G''$ if and only if $G$ is a disjoint union of the three types of connected components named earlier. Since $G \cong G'$ holds for such graphs as well, the problem statement follows right away.