In general, the integral $I$ does not converge to $0$ as $R \to \infty$:
For $R \notin \{(2k+1) \cdot \pi; k \in \mathbb{Z}\}$ let $$I(R) := \int_C \frac{e^{a \cdot z}}{1+e^{z}} \, dz$$ where the contour $C$ is parametrized by $z(t) = \imath \, R +t$, $t \in [-R,R]$. By definition, we have
$$\begin{align*} I(R) &= \imath \, e^{\imath \, a \cdot R} \cdot \int_{-R}^R \frac{e^{a \cdot t}}{1+e^{\imath \, R+t}} \, dt \\ &=\imath \, e^{\imath \, a \cdot R} \cdot \int_{-R}^R \frac{e^{a \cdot t}}{1+e^{\imath \, R+t}} \cdot \frac{1+e^{-\imath \, R + t}}{1+e^{-\imath \, R+t}} \, dt \\
&= \imath \, e^{\imath \, a \cdot R} \cdot \int_{-R}^R \frac{e^{a \cdot t}+e^{(a+1) \cdot t} \cdot e^{-\imath \, R}}{e^{2t}+\cos^2(R)+2e^t \cdot \cos^2(R)} \, dt \\
&= \imath \, e^{\imath \, a \cdot R} \cdot (I_1(R)-\imath \, I_2(R)) \end{align*}$$
where
$$\begin{align*} I_1(R) &:= \int_{-R}^R \frac{e^{a \cdot t}+e^{(a+1) \cdot t} \cdot \cos(R)}{e^{2t}+\cos^2(R)+2e^t \cdot \cos^2(R)} \, dt \\ I_2(R) &:= \sin(R) \cdot \int_{-R}^R \frac{e^{(a+1) \cdot t}}{e^{2t}+\cos^2(R)+2e^t \cdot \cos^2(R)} \, dt \end{align*}$$
Now pick a sequence $(R_n)_{n \in \mathbb{N}}$ such that $R_n \to \infty$ and $R_n \notin \{(2k+1) \cdot \pi; k \in \mathbb{N}\}$ for each $n \in \mathbb{N}$. The calculation above shows that $$\lim_{n \to \infty} I(R_n) = 0 \Leftrightarrow \lim_{n \to \infty} I_1(R_n) = \lim_{n \to \infty} I_2(R_n) = 0$$ First of all, by the positivity of the integrand,
$$\begin{align*} |I_2(R_n)| &\geq |\sin(R_n)| \cdot \underbrace{\int_{-R_n}^{R_n} \frac{e^{(a+1) \cdot t}}{(e^t+1)^2} \, dt}_{\geq c>0} \end{align*}$$
Consequently, $|\sin(R_n)| \to 0$ is a necessary condition for the convergence of $I_2(R_n)$. This implies $|\cos(R_n)| \to 1$. Without loss of generality, we may assume $\cos(R_n) \to 1$ or $\cos(R_n) \to -1$ (otherwise we choose a suitable subsequence). In the first case we find by the dominated convergence theorem,
$$I_2(R_n) \to I_2 := \int_{-\infty}^{\infty} \frac{e^{a \cdot t} + e^{(a+1) \cdot t}}{(e^t+1)^2} \, dt$$
Obviously, $I_2>0$, i.e. $I(R_n)$ does not converge to $0$. In the other case, i.e. $\cos(R_n) \to -1$, we have
$$I_2(R_n) \to I_2 := \int_{-\infty}^{\infty} \frac{e^{a \cdot t} - e^{(a+1) \cdot t}}{(e^t+1)^2} \, dt$$
Actually, it depends on $a$ whether this integral equals $0$ or not. For example, if $a=1/2$, it's not difficult to see that $I_2=0$. But in general, $I_2 \neq 0$. This means that in both cases $I(R_n) \to 0$ does in general not hold.
Concerning your original problem: Choose a sequence $R_n \to \infty$ and $R_n \notin \{(2k+1) \cdot \pi; k \in \mathbb{Z}\}$. Set $r_n := \frac{1}{\sqrt{R_n}}$. We consider the contour integral
$$\int_{D_n} \frac{e^{a \cdot z}}{1+e^z} \, dz$$
where the parametrization of $D_n$ is given by $z(t) = R_n \cdot e^{\imath \, t}$, $t \in [0,\pi]$. We split up the integral as follows
$$\begin{align*} \int_{D_n} \frac{e^{a \cdot z}}{1+e^z} \, dz &= \underbrace{\int_0^{\frac{\pi}{2}-r_n} \frac{e^{a \cdot z(t)}}{1+e^{z(t)}} \cdot z'(t)\, dt}_{=: J_1} \\ &\quad + \underbrace{\int_{\frac{\pi}{2}-r_n}^{\frac{\pi}{2}+r_n} \frac{e^{a \cdot z(t)}}{1+e^{z(t)}} \cdot z'(t)\, dt}_{=: J_2}+\underbrace{\int_{\frac{\pi}{2}+r_n}^{\pi} \frac{e^{a \cdot z(t)}}{1+e^{z(t)}} \cdot z'(t)\, dt}_{=: J_3} \end{align*}$$
and estimate the terms separately. Some standard calculations and estimats yield
$$\begin{align*} |J_1| &\leq R_n \cdot \int_{0}^{\frac{\pi}{2}-r_n} \frac{e^{a \cdot R_n \cdot \cos(t)}}{e^{R_n \cdot \cos(t)}-1} \, dt \end{align*}$$
The mapping $t \mapsto \frac{e^{a \cdot R \cdot \cos(t)}}{e^{R \cdot \cos(t)}-1}$ is monotonely increasing on $[0,\pi/2-r_n]$, therefore
$$|J_1| \leq R_n \cdot \frac{\pi}{2} \cdot \frac{e^{a \cdot R_n \cdot \cos(\pi/2-r_n)}}{e^{R_n \cdot \cos(\pi/2-r_n)}-1} \to 0 \quad (n \to \infty)$$
(Note that $R_n \cdot \cos(\pi/2-r_n) \to \infty$ as $n \to \infty$ and $a \in (0,1)$.) Similarly, we find
$$|J_3| \leq R_n \cdot \frac{\pi}{2} \cdot \frac{e^{a \cdot R_n \cdot \cos(\pi/2+r_n)}}{1-e^{R_n \cdot \cos(\pi/2+r_n)}} \to 0 \quad (n \to \infty)$$
The convergence of the term $|J_2|$ is rather obvious. This shows
$$\left| \int_{D_n} \frac{e^{a \cdot z}}{1+e^z} \, dz \right| \to 0$$
as $n \to \infty$. Therefore, the claim follows from the residue theorem.
