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.

Addendum: I see that this example also appears in the paper "Criteria of nilpotency and influence of contranormal subgroups on the structure of infinite groups" (Turk. J. Math. 33 (2009), pp. 227-237) by L. A. Kurdachenko, J. Otal, and I. Y. Subbotin (page 235). Moreover, in this paper and their subsequent paper "On influence of contranormal subgroups on the structure of infinite groups" (Comm. Algebra 37 (2010), pp. 4542-4557), the authors describe several other classes of groups for which the condition discussed here is equivalent to nilpotency. For example, in the latter paper, they show that this holds for any group that is an extension of a solvable minimax group by a nilpotent group (Corollary 2.15).

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.

Addendum: I see that this example also appears in the paper "Criteria of nilpotency and influence of contranormal subgroups on the structure of infinite groups" (Turk. J. Math. 33 (2009), pp. 227-237) by L. A. Kurdachenko, J. Otal, and I. Y. Subbotin (page 235). Moreover, in this paper and their subsequent paper "On influence of contranormal subgroups on the structure of infinite groups" (Comm. Algebra 37 (2010), pp. 4542-4557), the authors describe several other classes of groups for which the condition discussed here is equivalent to nilpotency. For example, in the latter paper, they show that this holds for any group that is an extension of a solvable minimax group by a nilpotent group (Corollary 2.15).

Other examples answering the question (as well as further results in the same vein as the two papers mentioned above) may be found in the paper "Groups with no proper contranormal subgroups"(Publ. Math. 64 (2020), pp. 183-194) by B.A.F. Wehrfritz.

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.

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.

Addendum: I see that this example also appears in the paper "Criteria of nilpotency and influence of contranormal subgroups on the structure of infinite groups" (Turk. J. Math. 33 (2009), pp. 227-237) by L. A. Kurdachenko, J. Otal, and I. Y. Subbotin (page 235). Moreover, in this paper and their subsequent paper "On influence of contranormal subgroups on the structure of infinite groups" (Comm. Algebra 37 (2010), pp. 4542-4557), the authors describe several other classes of groups for which the condition discussed here is equivalent to nilpotency. For example, in the latter paper, they show that this holds for any group that is an extension of a solvable minimax group by a nilpotent group (Corollary 2.15).