Here is an example of a metabelian group with your property in which not every subgroup is subnormal. Let $p$ be a prime and $\hat{\mathbb Z}_p$ denote the ring of $p$-adic integers. Let $Q$ be the multiplicative group consisting of all the $p$-adic integers that are congruent to 1 modulo $p$. Set $G:=\hat{\mathbb Z}_p\rtimes~Q$ using the action of $Q$ on $\hat{\mathbb Z}_p$ by multiplication.
Let $\epsilon:G\to Q$ be the canonical epimorphism that arises from the semidirect product structure, and let $A\cong \hat{\mathbb Z}_p$ be the kernel of $\epsilon$. It will be convenient to write $A$ multiplicatively. Hence, taking $t$ to be the element of $A$ corresponding to $1\in \hat{\mathbb Z}_p$, we will view $A$ as the set $\{t^\xi\ :\ \xi\in \hat{\mathbb Z}_p\}$.
To establish your property, let $H$ be a subgroup of $G$ such that $H^G=G$. Then $\epsilon(H)=Q$, and it will therefore follow that $H=G$ if we can verify that $t\in H\cap A$. To show this, observe that, for any $h\in H$ and $a\in A$, we have $h\, a\, h^{-1}=a^{\epsilon(h)}$, so that $a^{-1}\, h\, a=a^{\epsilon(h)-1}\, h$. Thus any product of elements of the form $a^{-1}\, h\, a$ with $a\in A$ and $h\in H$ is contained in $A^pH$. As a result, $A=H^G\cap A=A^p(H\cap A)$, which implies that $H\cap A$ must contain an element of the form $t^\xi$ for some $\xi\in Q$. Conjugating by an element of $H$ whose image under $\epsilon$ is $\xi^{-1}$ yields that $t\in H\cap A$. Therefore $H=G$.
This example was constructed together with Sezgin Sezer, who communicated your question to me.
Note that every finite group with your property must be nilpotent. As a consequence, there are also no nonnilpotent finitely generated solvable groups with the property. This is due to a result of Derek Robinson that a finitely generated solvable group is nilpotent if and only if all its finite quotients are nilpotent.