I have some partial answers.
- It is not hard to construct a Dirichlet series $$f(z)=\sum_{n=1}^\infty a_ne^{\lambda_n z}$$ which converges to $0$ absolutely and uniformly on the real line but does not converge in the complex plane. It is constructed as a sum of 3 series $f=f_0+f_1+f_2.$ Let $f_1$ be a series with imaginary exponents $\lambda_n$ which converges to an entire function in the closed lower half-plane, but not in the whole plane. Such series is not difficult to construct, see V. Bernstein, page 34, (see the full reference below) and there are simpler examples, with ordinary Dirichlet series. Then put $f_2=\overline{f_1(\overline{z})}$, and $f_0=-f_1-f_2$. So all three functions are entire. Now, according to Leontiev, EVERY entire function can be represented by a Dirichlet series which converges in the whole plane. Thus we have a Dirichlet series $f_0+f_1+f_2$ which converges on the real line to $0$ but does not converge in the plane.
A counterexample to the original question also requires real coefficients, this I do not know how to do (for $f_0$).
- It is clear from the work of Leontiev, that to obtain a reasonable theory, one has to restrict to exponents of finite upper density, $n=O(|\lambda_n|)$, otherwise there is no uniqueness in $C$. In the result I cited above the expansion of $f_0$ is highly non-unique.
But there is one more reasonable condition, that the $\lambda_n$ cannot come too close together. This is called "finite index of concentration" in Bernstein's book. When you have a finite index of concentration, $f_1$ and $f_2$ as above do not exist: if a Dirichlet series (with say imaginary exponents) converges to an entire function in a half-plane, then it converges everywhere.
- Assuming finite upper density AND finite index of concentration, I believe that I can prove $R$-linear independence. In fact a stronger property: if a Dirichlet series (with complex coefficients) converges on $R$ to $0$, then it has all zero coefficients.
The idea of the proof is following. Split the exponents into three sets. a) those in $|\arg(z-\pi/2)|<\pi/10,$ b) those in $|\arg(z+\pi/2)|<\pi/10,$ and the rest. So the series is split as $f_1+f_2+f_3$.
Now it is easy to prove that if $f_3$ converges at all real points, then it converges uniformly and abcolutely on all compacts in the plane. And thus represents an entire function. It is also easy to prove that for $f_1,f_2$ convergence at all real points implies uniform and absolute convergence in the closed upper and lower half-planes, respectively. So we have 3 functions, one entire one analytic in the lower halfplane and continuous on the closed lower half-plane, and the last one is analytic in the upper half-plane, continuous in the closure. And on the real line their sum is 0.
It follows (from the removable singularity theorem) that all three are entire. If the index of concentration is finite this would imply that the series of $f_1$ and $f_2$ converge in the whole plane. By uniququeness the sum is zero in the whole plane and we are done.
This is only a sketch because the statement that finite index of concentration implies convergence in the whole plane is proved (by Bernstein) only for imaginary exponents. However I believe that I can extend this to the case of exponents in a small sector around the imaginary axis.
Vladimir Bernstein's book is "Lecons sur les progress recent de la theorie des series de Dirichlet", Paris 1933. This is the most comprehensive book on Dirichlet series, but unfortunately only with real exponents.
The application to the functional equation mentioned by the author of the problem is not a good justification for the study of the problem in such generality. The set of exponentials there is very simple, and certainly we have $R$-linear independence for SUCH set of exponentials. Besides the theorem stated as an application has been proved in an elementary way.