Regaridng to the question Real numbers which are writable as a differences of two transcendental numbers now, it is important to know answer of the following question:
Is it true that for every real number $x\neq 0$ there exist linearly independent transcendental numbers $\alpha$ and $\beta$ (i.e., $r\alpha+s\beta=0$ implies $r=s=0$, for every rational numbers $r,s$) such that $x=\alpha-\beta$?
(note that it is true if $x$ is an algebraic number).
I will follow the idea of this answer by barak manos.
If $x \neq 0$ is algebraic, you can take the transcendental numbers $a=x+\pi$ and $b=-\pi$, which are $\Bbb Q$-linearly independent.
Indeed, if $r(x+\pi)-s\pi = 0$ (with $r,s \in \Bbb Q$) and $r \neq 0$ then $x=(s/r - 1)\pi$. Since $x \neq 0$, we know that $s/r-1 \neq 0$ and then $x$ would be transcendental ; contraditction. Therefore $r=0$, and then $s=0$.
If $x$ is transcendental, you can take the transcendental numbers $a=\sqrt{2}x$ and $b=(\sqrt{2}-1)x$, which are $\Bbb Q$-linearly independent.
Indeed, if $r\sqrt{2}x+s(\sqrt{2}-1)x=0$ (with $r,s \in \Bbb Q$), then $\sqrt 2 (r+s)=s$, so that $r=-s$ (otherwise $\sqrt 2$ would be rational...), and then $s=0$, which leads to $r=0$.