Here's a nice example of two extensions of $G$ by $H$ that are not equivalent, but with isomorphic middle groups. In fact, one of the extensions splits and the other does not.
Let $A=\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$, let $B=\mathbb{Z}/p\mathbb{Z}$; let $b$ be a generator of $B$. Let $B$ act on $A$ by letting $(a_1,a_2)^{b^r} = (a_1,a_2+ra_1)$.
Let $E=(A\rtimes B)\times (\mathbb{Z}/p\mathbb{Z})$. Note that $A$ is a normal subgroup of $E$, and that $E$ can be realized as a semidirect product of $A$ by $B\times(\mathbb{Z}/p\mathbb{Z})$. Thus, we have a split exact sequence $1 \longrightarrow \frac{\mathbb{Z}}{p\mathbb{Z}}\times\frac{\mathbb{Z}}{p\mathbb{Z}} \longrightarrow E \longrightarrow \frac{\mathbb{Z}}{p\mathbb{Z}}\times\frac{\mathbb{Z}}{p\mathbb{Z}}\longrightarrow 1.$
On the other hand, let $D$ be the subgroup of $E$ generated by the second direct factor of $A$ (which is central in $A\rtimes B$, hence in $E$) and the direct factor $\mathbb{Z}/p\mathbb{Z}$. This subgroup is isomorphic to $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$, and is central in $E$. Thus, we have an exact sequence $1\longrightarrow \frac{\mathbb{Z}}{p\mathbb{Z}}\times\frac{\mathbb{Z}}{p\mathbb{Z}} \longrightarrow E \longrightarrow \frac{\mathbb{Z}}{p\mathbb{Z}}\times\frac{\mathbb{Z}}{p\mathbb{Z}}\longrightarrow 1$ induced by this subgroup. The extension cannot be split, because the subgroup $D$ is central, so if the extension were split it would be a direct product (which it is not). So we have two extensions of $(\mathbb{Z}/p\mathbb{Z})\times(\mathbb{Z}/p\mathbb{Z})$ by itself, which yield isomorphic groups, but one of the extensions splits and the other does not, so the two extensions are not equivalent.
As to why we use this definition: for extensions, we are actually interested in how we are building the extension, and not only on the resulting groups. The definition of equivalence is such that it captures the idea that the two groups are built up from $G$ and $H$ in "essentially the same way", which of course implies isomorphisms. But as the example above shows, we can end up with the same kind of group by considering essentially different ways of having $H$ act on $G$. It's certainly interesting to know that the resulting groups are isomorphic even though the extensions are not equivalent, but we don't want the extensions to be equivalent because we do care about how we arrived at the group $E$, not just that we arrived.