Let me develop your idea of using two concentric contours:
Let $0 < \delta < R$ and consider the $C = C(\delta, R)$ consisting of
- $\Gamma = \Gamma(R)$ : upper semi-circular arc of radius $R$ centered at $0$,
- $\gamma = \gamma_{\delta}$ : upper semi-circular arc of radius $\delta$ centered at $0$,
- $L = L(\delta, R) = [-R, -\delta] \cup [\delta, R]$.
Of course, we impose counter-clockwise orientation to $C$ as usual. Also we inherit orientation of $C$ to each components $\Gamma$, $\gamma$, $L$. Now let $b > 0$ and consider
$$ f(z) = \frac{\exp( a e^{ibz} )}{z}. $$
Since $f$ has no pole in the interior of $C$, we have
$$ 0
= \oint_{C} f(z) \, dz
= \int_{\Gamma} f(z) \, dz + \int_{\gamma} f(z) \, dz + \int_{L} f(z) \, dz. $$
We will take $\delta \to 0^+$ and $R \to \infty$. To this end, we make the following observations:
Parametrize $\Gamma$ by $z = R e^{i\theta}$. Then
$$ \int_{\Gamma} f(z) \, dz
= i \int_{0}^{\pi} \exp\left( a e^{-bR\sin\theta}(\cos(bR\cos\theta)) + i\sin(bR\cos\theta)) \right) \, d\theta. $$
Since the integrand of the RHS is bounded by $e^a$, by the dominated convergence theorem we can pass limit into the integral as $R\to\infty$. The result is
$$ \lim_{R\to\infty} \int_{\Gamma(R)} f(z) \, dz = i\pi.$$
Similar computation for $\gamma$ shows that
$$ \lim_{\delta\to0^+} \int_{\gamma(\delta)} f(z) \, dz = -i\pi e^a.$$
The integral of $f$ along $L$ is related to our integral by
\begin{align*}
\int_{\delta}^{R} e^{a\cos (bx)}\sin(a\sin (bx)) \, \frac{dx}{x}
&= \frac{1}{2}\operatorname{Im} \int_{L(\delta, R)} f(z) \, dz \\
&= \frac{1}{2}\operatorname{Im} \left( -\int_{\Gamma(R)} f(z) \, dz - \int_{\gamma(\delta)} f(z) \, dz \right).
\end{align*}
Therefore, taking $\delta \to 0$ and $R\to\infty$ we obtain
$$ \int_{0}^{\infty} e^{a\cos (bx)}\sin(a\sin (bx)) \, \frac{dx}{x}
= \frac{\pi}{2}(e^a - 1) $$
as expected.