This is probably overkill, but anyway:
First, note that if $G$ is abelian and torsion free, then an equation of the form $nx = a$ with $n\gt 0$ has at most one solution. For if $g$ and $h$ are both solutions, then $ng = a = nh$, hence $0 = a-a = ng-nh = n(g-h)$, and since $n\gt 0$, that means that $g-h$ is a torsion element. But since $G$ is torsion free, this means $g-h = 0$, hence $g=h$. Thus, the equation has at most one solution.
Now, for the rest and to give context... To recall the definitions:
Theorem. If $G$ is a torsion free abelian group, then any two maximal independent sets in $G$ have the same cardinality. If a maximal independent subset of $G$ has $r$ elements, then $G$ is an additive subgroup of an $r$-dimensional vector space over $\mathbb{Q}$.
Proof. We can embedd $G$ into a divisible group; then moding out by the torsion of the divisible group, we obtain an embedding of $G$ into a torsion-free divisible group $D$, which has a natural structure of a $\mathbb{Q}$-vector space (since $nx=y$ has a unique solution for any $n\gt 0$). Now let $X$ be a maximal independent subset of $G$. Then the subspace generated by $X$, viewing $X$ as a subset of the $\mathbb{Q}$-vector space $D$, contains $G$ (since for any $g\in G$ there exists $n\in \mathbb{Z}$, $n\neq 0$, such that $ng$ is in the subgroup generated by $X$); so $G$ can be embedded in a vector space of dimension $|X|$). But that means that $X$ is a basis for the subspace generated by $G$, so the cardinality of $X$ is completely determined by $G$. QED
The rank of a torsion-free abelian group $G$ is the number of elements in a maximal independent subset.
If $G$ is a torsion-free abelian group, $x\in G$, and $p$ is a prime, then the $p$-height of $x$, $h_p(x)$, is the highest power of $p$ "dividing $x$ in $G$"; that is, the largest $n$ such that $p^ng = x$ has a solution in $G$; if the equation is solvable for all $n$, then $h_p(x) = \infty$.
If $x\neq 0$ is an element of a torsion-free group $G$, then the height sequence of $x$ is the sequence
$$h(x) = (h_2(x), h_3(x), h_5(x),\ldots,h_p(x),\ldots).$$
A characteristic is a sequence of nonnegative integers and the symbol $\infty$. Two characteristics are equivalent if and only if they have $\infty$ in the same coordinates and they differ in at most a finite number of other coordinates. An equivalence class of characteristics is called a type.
Lemma. If $G$ is a torsion free group of rank $1$, and $x,y\in G$ are both nonzero, then their height sequences are equivalent.
Proof. If $y=nx$, with $n=p_1^{e_1}\cdots p_t^{e_t}$, then trivially, $h_p(y)\geq h_p(x)$ in all cases, with equality for certain if $h_p(x)=\infty$ for all $p$.
For $p\neq p_i$, $i=1,\ldots,t$, suppose that $p^kg = y = nx$. Then $\gcd(p,n)=1$, so there exists $r$ and $s$ such that $p^kr + sn =1$. Hence
$$x = (p^kr + sn)x = p^k(rx) + s(nx) = p^k(rx) + sy = p^k(rx)+s(p^kg) = p^k(rx+sg),$$
so $h_p(x)\geq h_p(y)$.
Finally, $h_{p_i}(y) = e_i + h_{p_i}(x)$ (essentially the same argument), with the convention that $e_i+\infty=\infty$. Thus, the two sequences are equivalent.
In the general case, there are nonzero integers $m$ and $n$ such that $my=nx$. From the previous part, we know the height sequences for $y$ and $my$ are equivalent; the height sequences for $x$ and $nx$ are equivalent; and therefore the height sequences for $y$ and $x$ are equivalent. QED
In light of this, if $G$ is a torsion free abelian group of rank $1$, then we can define the type of $G$ to be the equivalence class of the height sequence of any nonzero element of $G$.
Theorem. Let $G$ and $G'$ be torsion-free abelian groups of rank $1$. Then $G\cong G'$ if and only if the type of $G$ is equal to the type of $G'$.
Proof. If $\phi\colon G\to G'$ is any homomorphism, and $x\in G$ is not in the kernel, then $h_p(x)\leq h_p(\phi(x))$ for all primes $p$; if $\phi$ is invertible, then we get the reverse inequality by applying $\phi^{-1}$, so $G\cong G'$ implies the types are equal.
Conversely, if $x\in G$ and $x'\in G'$, then $x$ and $x'$ have equivalent height sequences. Let $P$ be the finite (and possibly empty) set of primes for which $h_p(x)\lt h_p(x')$, let $Q$ be the finite (and possibly empty) set of primes for which $h_p(x)\gt h_p(x')$. Let
$$m = \prod_{p\in P} p^{h_p(x')-h_p(x)},\qquad n=\prod_{p\in Q} p^{h_p(x)-h_p(x')}.$$
Then $mx$ and $nx'$ have the same height sequence (using calculations as we did above). Thus, we can find elements of $G$ and $G'$ that have identical height sequences.
Since both $G$ and $G'$ are torsion free of rank $1$, we can view them as subgroups of $\mathbb{Q}$, and we have elements $y\in G$ and $y'\in G'$ that have the same height sequence, both nonzero. Write $y = \frac{a}{b}$ and $y'=\frac{a'}{b'}$ (viewing them as rational numbers). Then $\frac{b}{a}G$ is isomorphic to $\frac{b'}{a'}G'$, both contain $1$, and $1$ has the same height sequence in both.
It now follows that $\frac{b}{a}G = \frac{b'}{a'}G'$: for any $p$ for which $h_p(1)=\infty$, the groups contain the corresponding Prüfer group; if $h_p(1)=e_p\lt\infty$, then it contains $\frac{1}{p^{e_p}}$ and no reciprocal of any higher power of $p$; and that is it. QED
Since the height sequence of $ta$ is completely determined by the height sequence of $a$ and the prime factorization of $t$, your question comes down to proving the following:
If $G$ is a torsion free abelian group of rank $1$, $a$ and $b$ are two nonzero elements, and the height sequences of $a$ and $b$ are identical, then $nx = a$ has a solution if and only if $nx=b$ has a solution.
This follows from:
Proposition Let $G$ be a torsion free abelian group of rank $1$, let $a\in G$ be a nonzero element, let $h(a) = (h_2(a),h_3(a),\ldots,)$ be the height sequence for $a$, and let $n= p_1^{e_1}\cdots p_k^{e_k}$ be a positive integer. Then $nx=a$ has a solution if and only if $e_i\leq h_{p_i}(a)$ for each $i$.
(I'll let you prove this one on your own; one possibility for the "if" part is to do induction on $k$.)
The highlighted question now has an immediate answer, since whether or not $nx=y$ has a solution depends only on the height sequence of $y$ and the prime factorization of $n$, and since $a$ and $b$ have identical height sequences it follows that $nx=a$ and $nx=b$ either both have solutions, or neither has solutions; and if it has solutions, it has at most one since the group is torsion free.
Are you used to Zorn's lemma proofs? Because here is a quick one:
Let $A$ be a torsion-free abelian group. A sub-monoid of $A$ is a subset $P$ containing $0$ and closed under addition. We'll say that $P$ is pointed if it does not contain both $x$ and $-x$ for any nonzero $x \in A$. Clearly, the union of an ascending collection of pointed sub-monoids is itself a pointed sub-monoid. So, by Zorn's lemma, there is a maximal element pointed sub-monoid of $A$.
Claim 1: For any $x \in A$, and any positive integer $k$, if $kx \in P$ then $x \in P$.
Proof: Let $Q = \{ x \in A : kx \in P\ \text{for some } k \in \mathbb{Z}_{>0} \}$. Note that $Q$ is a sub-monoid: It clearly contains $0$ and, if $k_1 x_1 \in P$ and $k_2 x_2 \in P$, then $k_1 k_2 (x_1+x_2) \in P$. We also claim that $Q$ is pointed: If there is $x \neq 0$ such that $k x \in P$ and $\ell(-x) \in P$ then $k \ell x \in P$ and $- k \ell x \in P$. Since $A$ is torsion-free, $k \ell x \neq 0$, contradicting that $P$ is pointed.
So $Q$ is a pointed monoid containing $P$ and, by the maximality of $P$, we have $Q=P$. In other words, if $kx \in P$ for $k>0$, then $x \in P$, as desired. $\square$
Claim 2: For any $x$ in $A$, either $x$ or $-x \in P$.
Proof: Suppose that $-x \not \in P$.
Let $P' = \{ y+nx : y \in P, n \in \mathbb{Z}_{\geq 0} \}$. Clearly $P'$ is a sub-monoid of $A$. We claim that $P'$ is pointed. If not, then we have $y_1 + n_1 x = - (y_2 + n_2 x)$ for some $y_1$, $y_2 \in A$ and some $n_1$, $n_2 \geq 0$ with $y_1+n_1 x \neq 0$.
So $y_1 + y_2 =- (n_1+n_2) x$. If $n_1=n_2=0$, then $y_1 = - y_2$ and, since $P$ is pointed, $y_1=y_2=0$. In this case, $y_1+n_1 x = 0$, while we assumed otherwise.
Alternatively, suppose that $n_1+n_2>0$. We know $y_1 +y_2 \in P$ and by claim 1, the identity $y_1 + y_2 =- (n_1+n_2) x$ then forces $-x \in P$, a contradiction.
So we now know that $P'$ is pointed. So $P'=P$ and $x \in P$, as desired. $\square$
With $P$ as above, define $y \leq z$ if $z-y \in P$. This is reflexive (as $0 \in P$) and transitive (as $P$ is closed under addition.) It is anti-symmetric (because $P$ is pointed) and a total order by the claim.
So we have equipped $A$ with a linear order. As $P$ is closed under addition, the order is compatible with the group structure, meaning that $w \leq x$ and $y \leq z$ means $w+y \leq x+z$.
Best Answer
As David Wheeler says in the comments, we cannot generally speak of the torsion subgroup of groups that are not abelian, because the torsion points need not be closed under multiplication; an obvious example being the infinite dihedral group. On the other hand the proof you've written does demonstrate that all torsion points are contained in $H$.
It is possible for non-abelian groups to have torsion subgroups, of course. If $G$ is finite then the torsion points are just all of $G$, for instance, and Wikipedia states that the torsion subset of any nilpotent group forms a normal subgroup (so this would include infinite groups). The orders of elements do not behave predictably in nonabelian settings, however: whereas in abelian groups we have that the order of $ab$ is the $\rm lcm$ of the orders of $a$ and $b$, we can actually pick any integers $r,s,t$ and there will be a group $G$ with elements $a,b,ab$ of those orders respectively (this is Thm 1.64 in Milne's freely available group theory course notes), specifically the $\rm PSL_2$s of some finite fields.