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.
Here is a fairly simple argument directly from the universal property. Given an abelian group $A$, define a group $F$ of "fractions" $\frac{a}{n}$ where $a\in A$ and $n\in\mathbb{Z}\setminus\{0\}$. We impose an equivalence relation on these fractions by saying $\frac{a}{n}=\frac{b}{m}$ iff there exists $k\in\mathbb{Z}\setminus\{0\}$ such that $k(ma-nb)=0$. Using the usual formula for addition of fractions, you can check that the set $F$ of equivalence classes of fractions forms an abelian group.
There is now a bilinear map $f:A\times\mathbb{Q}\to F$ defined by $f(a,m/n)=\frac{am}{n}$. This induces a homomorphism $g:A\otimes\mathbb{Q}\to F$ which sends $a\otimes 1$ to the fraction $\frac{a}{1}$. So to show that $a\otimes 1\neq 0$, it suffices to show the fraction $\frac{a}{1}$ is nonzero. The zero element of $F$ is the fraction $\frac{0}{1}$, and $\frac{a}{1}=\frac{0}{1}$ iff there exists $k\in\mathbb{Z}\setminus\{0\}$ such that $k(1\cdot a-1\cdot 0)=0$, or in other words such that $ka=0$. So if $a$ is a non-torsion element, then $a\otimes 1\neq 0$. (In fact, this map $g$ is actually an isomorphism, but that takes more work to prove.)
The case of $\mathbb{R}$ follows from the case of $\mathbb{Q}$ since $\mathbb{Q}$ is a direct summand of $\mathbb{R}$ and tensor products distribute over direct sums.
From a broader perspective, what's really going on here is that $\mathbb{Q}$ (or $\mathbb{R}$) is a flat $\mathbb{Z}$-module, meaning that if $A$ is a $\mathbb{Z}$-module and $B\subseteq A$ is a submodule, then the map $B\otimes\mathbb{Q}\to A\otimes\mathbb{Q}$ induced by the bilinear map sending $(b,q)\in\mathbb{Q}$ to $b\otimes q\in A\otimes\mathbb{Q}$ is injective. So, in particular, if $a\in A$ is any element such that $na\neq 0$ for all $n$, the submodule $B\subseteq A$ generated by $a$ is isomorphic to $\mathbb{Z}$. Since $\mathbb{Z}\otimes\mathbb{Q}\cong\mathbb{Q}$ and we have an injective map $\mathbb{Z}\otimes\mathbb{Q}\cong B\otimes\mathbb{Q}\to A\otimes\mathbb{Q}$, this implies $A\otimes\mathbb{Q}$ is nontrivial (and in fact $a\otimes1\neq 0$, since it is the image of the nonzero element $1\otimes 1\in \mathbb{Z}\otimes\mathbb{Q}$).
The argument using fractions above can be generalized to show that if $R$ is any commutative ring, then any localization of $R$ is flat as an $R$-module. For $\mathbb{Z}$-modules, flatness can be characterized quite simply in general: a $\mathbb{Z}$-module is flat iff it is torsion-free. See Show that a Z-module A is flat if and only if it is torsion free? for some sketches of the proof.
Best Answer
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 $w+y \leq x+z$.