Is
$$\mathrm{U}_{\omega}=\Big\{x\mid\forall z\Big(\big(\emptyset\in z\wedge \forall u, v\;(u,v\in z\rightarrow\{w\mid w\in u\vee w=v\}\in z)\big)\rightarrow x\in z\Big)\Big\}$$
identical with the set $\mathrm{V}_{\omega}$ of hereditarily finite sets, i.e. the level $\omega$ of the cumulative hierarchy?
I assume that the definition of $U_\omega$ has a typo: adjunction operation usually means the following binary operation: $$u;v:=u\cup\{v\}$$
If you just assume $z$ in the definition of $U_\omega$ is closed under successor operator $u\mapsto u\cup\{u\}$, then $U_\omega$ would be $\omega$.
It suffices to show that every set which contains $\varnothing$ and is closed under adjunction contains $V_\omega$.
Let $z$ be a set such that $\varnothing\in z$ and is closed under adjunction. Assume inductively that $V_n\subseteq z$. For $x\in V_{n+1}=\mathcal{P}(V_n)$, we have $x=\{a_0,\cdots, a_k\}$ for some $a_0,\cdots,a_k\in V_n$. By induction on $i\le k$, you can see that $\{a_0,\cdots, a_i\}\in z$ (In the case of $i=0$, $\{a_0\}$ is the adjunction of $\varnothing$ and $a_0\in z$.) Hence $x\in z$. This shows $V_{n+1}\subseteq z$.
