The answer is no. Let us show something stronger: for every $u\neq 1$, the set of $n$ such that $u$ has an $n$-root is bounded. Let me show this for $$G=\mathrm{BS}(m,n)=\langle t,x\mid tx^mt^{-1}=x^n\rangle$$ whenever neither $m/n$ nor $m/n$ is an integer.
Indeed, $G$ acts on its Bass-Serre tree $T$, with a vertex $v_0$ with stabilizer $\langle x\rangle$. So every vertex $v$ has cyclic stabilizer, with a unique generator $\xi_v$ conjugate to $x$.
Let $u$ satisfy the previous property. This property forces $u$ to act as an elliptic element (indeed if $u$ were loxodromic of translation length $R$, it could not have $p$-roots for $p>R$).
The Bass-Serre tree has a unique height function $h$ satisfying: $h(tv)=h(v)+1$ and $h(xv)=h(v)$ for all vertices $v$. Then every vertex $v$ has exactly $n+m$ neighbors, consisting of two $G_v$-orbits: one orbit of $n$ neighbors $v'$ with $h(v')=h(v)+1$ (called right neighbors) and one orbit of $m$ neighbors $v'$ with $h(v')=h(v)-1$ (called left neighbors). If $v$ is a vertex and $v'$ is a right neighbor, then we have the formula $\xi_v^n=\xi_{v'}^m$.
Let $s\in G$ be an elliptic element (for the action on $T$). Let $T^s$ be the tree of fixed points of $s$. For every vertex $v\in T^s$, we have $s\in G_v$, so $s=\xi_v^{q_u(v)}$ for some unique $q_s(v)\in\mathbf{Z}$. If $v,v'\in T^s$ with $v'$ right neighbor of $v$, then by the previous formula, we have $q_s(v')=\frac{m}{n}q_s(v)$. Hence by connectedness of $T^s$, for any two vertices $v,v'\in T^s$ we have $q_s(v')=\left(\frac{m}{n}\right)^{h(v')-h(v)}q_s(v)$.
If neither $m/n$ nor $n/m$ is an integer and $s\neq 1$, it follows that $h(T^s)$ is bounded, and in turn, it follows that $q_s$ is bounded on $T^s$.
Now let $u\neq 1$ have roots of unbounded order. As we have seen, $u$ is elliptic. Then for some $R$ we have $|q_u(v)|\le R$ for all $v\in T^u$. Suppose that $u=s^N$. So $s$ is elliptic as well, so fixes a vertex $v$, and necessarily $v\in T^u$. It follows that $N\le R$.
Comment: if $n/m=\pm 1$, the same conclusion holds by a distinct (easier) argument, while if exactly one of $m/n$ or $n/m$ is an integer $r$ (so $|r|\ge 2$), the conclusion fails as one can then check that $\mathrm{BS}(m,n)$ then contains an isomorphic copy of $\mathbf{Z}[1/r]$: in $\mathrm{BS}(m,mr)$ we have $x^m=(t^{-p}x^mt^p)^{r^p}$ for all positive integers $p$.
Another fact (to address Derek's comments): let $I$ be an infinite set of positive integers. Write $G^n=\{g^n:g\in G\}$. Say that a group $G$ satisfies Property $P_I$ if $\bigcap_{n\in I} G^n=\{1\}$, and $g^n=1$ implies $g=1$ for every $n\in I$. Then Property $P_I$ passes to group extensions.
If $G$ is free, then it satisfies $P_I$ for every infinite $I$.
If $G=\mathbf{Z}[1/k]$, then it satisfies $P_I$ for every infinite $I$ of integers coprime to $k$.
Define $k=mn/\gcd(m,n)^2$. Then there is an obvious homomorphism $\mathrm{BS}(m,n)\to\mathbf{Z}[1/k]\rtimes_{n/m}\mathbf{Z}$. It follows that for every infinite $I$ of elements coprime to $k$, $\mathrm{BS}(m,n)$ has Property $P_I$. In particular, every nontrivial element is divisible by only finitely many primes.
Best Answer
"determine the normal closure" is not very clear. It seems you want to identify the isomorphism type of the normal closure.
Namely, this kernel $\mathrm{BS}_0(m,n)$ is an infinitely iterated amalgam $$\cdots\ast_\mathbf{Z}\mathbf{Z}\ast_\mathbf{Z}\mathbf{Z}\ast_\mathbf{Z}\mathbf{Z}\ast_\mathbf{Z}\cdots$$ where each left embedding is $k\mapsto nk$ and each right embedding is $k\mapsto mk$. This is a general fact about HNN extensions (description of the isomorphism type of the kernel of the canonical homomorphism from a given HNN-extension onto $\mathbf{Z}$), which can be found in Serre's book. In particular, this kernel $\mathrm{BS}_0(m,n)$ is not free as soon as $\max(|m|,|n|)\ge 2$. Indeed,
On the other hand, the kernel $\mathrm{BS}_{00}(m,n)$ of the canonical homomorphism onto $\mathbf{Z}[m/n,n/m]\rtimes_{m/n}\mathbf{Z}$ is free, since it acts freely on the Bass-Serre tree.
Note: $\mathrm{BS}_0(m,n)$ is locally residually finite. Also $\mathrm{BS}_0(m,n)$ is not residually finite if $|m|,|n|$ are coprime and $\ge 2$: this is due to R. Campbell (1990 Proc AMS). I don't know if it's also non-residually-finite whenever $2\le |m|<|n|$, for instance, $(m,n)=(2,4)$.
Edit: the free subgroup $\mathrm{BS}_{00}(m,n)$ is infinitely generated as soon as $2\le |m|<|n|$. Since $2\le \min(|m|,|n|)$, it is not trivial, hence contains a loxodromic element for the action on the Bass-Serre tree. Hence it has a unique minimal nonempty subtree (the convex hull of the union of axes of loxodromic elements) for the action on the Bass-Serre tree $T$ of the HNN extension defining $\mathrm{BS}(m,n)$; hence this subtree is invariant under the whole group, and hence by vertex-transitivity, this is the whole tree.
The tree $T$ naturally carries a Busemann-function (defined up to addition with an integral constant), so that every vertex $v$ has $n+m$ adjacent vertices: $m$ vertices $w$ with $b(w)=b(v)-1$ and $n$ vertices $w$ with $b(w)=b(v)+1$. Let $G$ is the group of automorphisms of $T$ preserving $b+\mathbf{Z}$, so the locally compact group $G$ acts cocompactly on $T$ and $\mathrm{BS}(m,n)$ acts through $G$.
Now assume by contradiction that $\mathrm{BS}_{00}(m,n)$ is finitely generated: then since it acts freely and minimally on the tree $T$, the action is cocompact. Hence $\mathrm{BS}_{00}(m,n)$ is a cocompact lattice in $G$. But $G$ is not unimodular, by an easy argument (using $|m|\neq |n|$). This is a contradiction.