The integrand $\mathrm{e}^z$ is holomorphic for $\vert z \vert \le 1$ (check that), therefore the integral vanishes by the Cauchy integral theorem (wiki).
Now let's look at it like you do:
$$ \begin{eqnarray}
\int_0^{2 \pi} \mathrm{e}^{\cos(\theta)} \cos(\theta + \sin(\theta)) \mathrm{d} \theta &=&
\int_0^{2 \pi} \frac{\mathrm{d}}{\mathrm{d} \theta} \left( \mathrm{e}^{\cos(\theta)} \sin(\sin(\theta)) \right) \mathrm{d} \theta = \left. \mathrm{e}^{\cos(\theta)} \sin(\sin(\theta)) \right|_0^{2\pi} = 0
\end{eqnarray}
$$
Indeed:
$$ \begin{eqnarray}
\mathrm{e}^{\cos(\theta)} \cos(\theta + \sin(\theta)) &=& \mathrm{e}^{\cos(\theta)} \cos(\theta) \cos(\sin(\theta)) - \mathrm{e}^{\cos(\theta)} \sin(\theta) \sin(\sin(\theta) \\
&=& \mathrm{e}^{\cos(\theta)} \cdot \frac{\mathrm{d} \sin(\sin(\theta))}{\mathrm{d} \theta} +
\frac{\mathrm{d} \mathrm{e}^{\cos(\theta)}}{\mathrm{d} \theta} \cdot \sin(\sin(\theta)) \\
&=& \frac{\mathrm{d}}{\mathrm{d} \theta} \left( \mathrm{e}^{\cos(\theta)} \sin(\sin(\theta)) \right)
\end{eqnarray}
$$