Is there a torsion-free abelian group $A$ such that $A \not \simeq A \oplus \Bbb Z \simeq A \oplus \Bbb Z \oplus \Bbb Z$ (as groups)?

Notice that $\Bbb Z$ is not cancellable, so $A \oplus \Bbb Z \simeq (A \oplus \Bbb Z) \oplus \Bbb Z$ doesn't imply that $A \simeq A \oplus \Bbb Z$. Combined with this question, such a group $A$ would possibly provide an answer to that question.

An example of torsion free abelian group $A$ such that $A$ is isomorphic to $A \oplus \mathbb{Z}^2$, but not to $A \oplus \mathbb{Z}$ was given there. The example was the additive group of bounded sequences of elements of $\mathbb{Z}[\sqrt{2}]$, i.e. $$A = \left\{ (x_n)_{n \geq 1} \subset \Bbb Z[\sqrt 2] \;\;:\;\; \exists C>0,\; \forall n \geq 1,\; |x_n| \leq C \right\}$$ I wasn't able to adapt this example in order to answer my question.