Are solutions of the following equation countable : $ \frac{a\exp(ix)}x + \frac{b \exp(iy)}y = c $? I would like to prove that:
For given non zero complex numbers $a,b$ and $c$, the set of positive real numbers $x>0$, $y>0$ satisfying the equation:
$$
 \frac{a\exp(ix)}x + \frac{b \exp(iy)}y = c
$$
is countable, where $i$ is the imaginary (complex) unit.

This is a very partial answer attempting to cover all the cases, with a proven case and other cases at the state of conjectures without proofs.
First, have a look at the figures at the bottom ; it suffices to know at present that the common points to the two spirals are in correspondence with solutions. The first figure has a finite number of intersection points, the second one, where one of the spiral passes through the center of the other one, has a - denumerable - infinite number of intersection points. The third one illustrates the case of a common center with, again a (denumerable) infinite number of intersection points.
Let us consider the curve (S) parameterized in this way:
$$\gamma: t \to \dfrac{1}{t}e^{it} \ \ \text{for} \ \ t>0.$$
It is an hyperbolic spiral * ((https://en.wikipedia.org/wiki/Hyperbolic_spiral) with equation $r\theta=1$), different from other classical families of spirals, Archimedean, logarithmic, and Cornu spiral.
Two extreme cases:


*

*If $t \to 0$, $\gamma(t) \to +\infty$, i.e., (S) possesses an asymptote which is a straight line parallel to the real axis. More precisely, by considering $\dfrac{1}{t}e^{it}=\dfrac{1}{t}(1+it-\dfrac{1}{2}t^2+\cdots)$, this asymptote has equation  $y=1$.

*If $t \to \infty$, $\gamma(t) \to 0$ ; the curve spirals around a limit point, the origin that will be called the "center" of the spiral.
The equation given in the question can be transformed into:
$$\tag{1}\gamma(x)=c'+b'\gamma(y) \ \ \text{with} \ \ b'=-b/a, \ \ c'=c/a.$$
We have thus to consider the intersection of two spirals, the one in the LHS which is $(S)$, the other in the RHS, let us name it $(S')$, which is an enlarged, rotated and translated version of $(S)$, the enlargement factor being $|b'|$, the rotation angle being arg($b'$), and the translation given by $c'$.
Thus, two cases can occur :
Case 1 : (illustrated by Figure 3) : if $c'=0$ [the spirals' centers coincide] : in this case, let $b'=re^{i\theta}$ (thus $r$ and $\theta$ are fixed quantities ; $r$ is assumed $\neq 1$). (1) can be written under the form:
$$\tag{2}\gamma(x)=b'\gamma(y) \ \iff \ \dfrac{1}{x}e^{ix}=re^{i\theta} \dfrac{1}{y}e^{iy}$$
Equating modulus and argument on both sides gives 
$$\tag{3} \ \iff \begin{cases}y=rx \\ x=\theta+y+k2\pi \end{cases}$$
where each $k \in \mathbb N$ is associated to a specific turn around the origin.
Now, consider the resulting equation:
$$\tag{4}x=\theta+rx+k2\pi \ \iff x=\dfrac{1}{1-r}(\theta+k2\pi)$$
(take $k \in \mathbb N$ if $r<1$ and $k \in -\mathbb N$ if $r>1$) giving a unique solution for each turn (because once $x$ is known, $y=rx$ is known). Thus there are a denumerable infinite number of solutions.
Case 2: $c' \neq 0$. There are two subcases:
Subcase 2.1: one of the spirals passes through the center of the other (illustrated by Figure 2).
Conjecture: there is an (denumerable) infinite - number of intersection points.
Subcase 2.2: no spiral passes through the center of the other (illustrated by figure 1). 
Conjecture : there is a finite number of intersection points.

Remark about subcase 2.1: the fact that for example spiral (S') passes through the origin can be transcribed into the constraint 
$$\exists y \in \mathbb R, \ c'-b'\dfrac{1}{y}e^{iy}=0  \ \iff \ \ d'y=e^{iy}  \ \iff \ \ yre^{i \theta}=e^{iy}$$
(by setting $d':=\tfrac{c'}{b'}:=re^{i \theta}$). Thus, we must have simultaneously $yr=1$ and $\theta = y + k 2 \pi$. Therefore, the constraint is to have $d'$ such that $|d'|=1/(arg(d')+k 2 \pi)$ (which is a family of spirals).

(Figure 1: Random case : at most a finite number of intersection points. Here $b'=1-i, \ \ c'=0.1+0.2i$.)

(Figure 2: Case where one of the spirals passes through the center of the other.)

(Figure 3: case where the spirals have a common center : the common points have been computed using formula (4).)


*

*[I'm indebted to Yves Daoust for pointing to me the name 'hyperbolic spiral" I didn't know]

