Do there exist irreducible elements in this domain? I asked this question on MSE. Here also I have the same motive in the question.
Let $D= \{\,a_1x^{r_1} + \cdots + a_n x^{r_n} \, \vert  \, a_i \in \mathbb{C} \text{ for } i= 1,2,\dots,n \text{ and each $r_i$ is a real number such that } 0 \le r_1 < r_2 < \cdots <r_n   \,\}$
Given non zero, non unit element $f(x) = a_1x^{r_1} + \cdots + a_n x^{r_n}$ in $D$, if all of $r_i$ are rationals then we can find an integer $s$ such that $f(x^s) \in \mathbb{C}[x]$ and we can split $f(x^s)$ into linear factors in $\mathbb{C}[x]$ by which it follows that there exist a complex number $b$ such that $x^{\frac1s} - b $ divides $f(x)$. Since $x^{\frac1s} - b$ can be written as product of elements which are not unit, $f(x)$ cannot be an irreducible. Moreover this argument shows that $f(x)$ cannot be written as finite product of irreducibles.
In general, the above strategy works if we can find an integer $s$ such that $sr_i \in \mathbb{Q}$ for all $i$ then $f(x)$ cannot be written as a finite product of irreducibles.
Difficulties come in simplest cases such as $g(x) = 1 + x^e + x^{\pi}$ in which I failed to deduce whether this is irreducible or not. Are there any methods to check the reducibility of an element in $D$? This $D$ is certainly not a factorization domain but the stronger statement would be to check if all the non-zero , non-unit elements are reducible?
 A: Fix $m\ge 0$. Let $a_1,\dots,a_m\in\mathbf{R}$ be linearly independent over $\mathbf{Q}$.
Claim The element $1+\sum_{i=1}^mx^{a_i}$ is irreducible in $D=\mathbf{C}[\mathbf{R}_{\ge 0}]$ iff for all integers $n_1,\dots n_m\ge 1$, the polynomial $1+\sum_{i=1}^mz_i^{n_i}$ is irreducible in the polynomial algebra $\mathbf{C}[z_1,\dots,z_m]\,(=\mathbf{C}[\mathbf{N}^m])$.
Proof: one direction is straightforward: if $1+\sum_{i=1}^m{z_i^{n_i}}$ is not irreducible in $\mathbf{C}[z_1,\dots,z_m]$ we mechanically inherit such a factorization in $D$.
Conversely, suppose that $1+\sum_{i=1}^mx^{a_i}$ is reducible. Then there exists a finitely generated subgroup $\Lambda$ of $\mathbf{R}$, containing all $a_i$, such that it is also reducible in $\mathbf{C}[\Lambda]$. Enlarging $\Lambda$ if necessary, we can suppose that $\Lambda$ has a basis $(b_1,\dots,b_k)$, with $k\ge m$, such that for every $i\le m$ we have $a_i=n_ib_i$ for some $n_i\ge 1$. We deduce that $1+\sum_{i=1}^mz_i^{n_i}$ is reducible in $\mathbf{C}[z_1^{\pm 1},\dots,z_k^{\pm 1}]$, and hence in $\mathbf{C}[z_1,\dots,z_k]$ as well.

Both sides are clearly false for $m=0,1$. For $m=2$ this is what I claimed in a comment: $1+x^a+x^b$ is irreducible in $D$ (for $(1,a,b)$ linearly independent over $\mathbf{Q}$) iff for all $n,m\ge 1$, $1+x^n+y^k$ is irreducible in $\mathbf{C}[x,y]$. (I'm pretty sure it is, and this ought to be very standard.)

Edit: for $m=2$ this is indeed irreducible. Indeed, if $1+x^n+y^k$ were reducible ($n,k\ge 1$), then $1+x^{nk}+y^{nk}$ would be reducible too. So we can suppose $n=k$. Then one can check smoothness of the zero locus of the Fermat curve $x^n+y^n+z^n=0$ (see for instance this MathSE answer). A hypersurface in $\mathbf{P}^2$ defined by a polynomial is always connected (consequence of Bézout). Hence smoothness implies irreducibility. (A more direct argument is certainly not hard to find). So $D$ indeed has many irreducible elements.
Irreducibility for $m\ge 3$ probably follows from the $m=2$ case.
