This is an observation which is too big to fit in comments. I claim that if there exists $M>0$ such that for all $n$, $|\Im \lambda_n| < M$, then the answer is yes ($\mathbb{R}$-independence implies $\mathbb{C}$-independence). In this case it's easy to see that uniform convergence on compact intervals of $\mathbb{R}$ implies uniform convergence on compact regions of $\mathbb{C}$.

To see this, write $\lambda_n=x_n+y_ni$ where $x_n, y_n$ are real; writes $s=c+di$. Then $$\sum |a_ie^{\lambda_i s}|=\sum |a_i|e^{x_nc-y_nd}$$ which can be compared to $$\sum |a_i| e^{x_n c}.$$

But then uniform convergence on compact regions of $\mathbb{C}$ implies that the limit function $f$ is holomorphic, and it vanishes identically on $\mathbb{R}$, so $f$ must be identically zero. But then applying $\mathbb{C}$-independence, we have that all the $a_i=0$.

This points in the direction of a counterexample when the $y_i$ are unbounded---if we let the $|y_i|$ tend to $\infty$ rapidly this argument fails dramatically.

EDIT: The following actually assumes uniform absolute convergence on compact regions, and thus addresses a slightly different problem, as Zen Harper points out in the comments.

This is an observation which is too big to fit in comments. I claim that if there exists $M>0$ such that for all $n$, $|\Im \lambda_n| < M$, then the answer is yes ($\mathbb{R}$-independence implies $\mathbb{C}$-independence). In this case it's easy to see that uniform convergence on compact intervals of $\mathbb{R}$ implies uniform convergence on compact regions of $\mathbb{C}$.

To see this, write $\lambda_n=x_n+y_ni$ where $x_n, y_n$ are real; writes $s=c+di$. Then $$\sum |a_ie^{\lambda_i s}|=\sum |a_i|e^{x_nc-y_nd}$$ which can be compared to $$\sum |a_i| e^{x_n c}.$$

But then uniform convergence on compact regions of $\mathbb{C}$ implies that the limit function $f$ is holomorphic, and it vanishes identically on $\mathbb{R}$, so $f$ must be identically zero. But then applying $\mathbb{C}$-independence, we have that all the $a_i=0$.

This points in the direction of a counterexample when the $y_i$ are unbounded---if we let the $|y_i|$ tend to $\infty$ rapidly this argument fails dramatically.

This is a partial solution. If there exists $M>0$ such that for all $n$, $|\Im \lambda_n| < M$, then the answer is yes. In this case it's easy to see that uniform convergence on compact intervals of $\mathbb{R}$ implies uniform convergence on compact regions of $\mathbb{C}$.

To see this, write $\lambda_n=x_n+y_ni$ where $x_n, y_n$ are real; writes $s=c+di$. Then $$\sum |a_ie^{\lambda_i s}|=\sum |a_i|e^{x_nc-y_nd}$$ which can be compared to $$\sum |a_i| e^{x_n c}.$$

But then uniform convergence on compact regions of $\mathbb{C}$ implies that the limit function $f$ is holomorphic, and it vanishes identically on $\mathbb{R}$, so $f$ must be identically zero. But then applying $\mathbb{C}$-independence, we have that all the $a_i=0$.

This points in the direction of a counterexample when the $y_i$ are unbounded---if we let the $|y_i|$ tend to $\infty$ rapidly this argument fails dramatically.

This is an observation which is too big to fit in comments. I claim that if there exists $M>0$ such that for all $n$, $|\Im \lambda_n| < M$, then the answer is yes ($\mathbb{R}$-independence implies $\mathbb{C}$-independence). In this case it's easy to see that uniform convergence on compact intervals of $\mathbb{R}$ implies uniform convergence on compact regions of $\mathbb{C}$.

To see this, write $\lambda_n=x_n+y_ni$ where $x_n, y_n$ are real; writes $s=c+di$. Then $$\sum |a_ie^{\lambda_i s}|=\sum |a_i|e^{x_nc-y_nd}$$ which can be compared to $$\sum |a_i| e^{x_n c}.$$

But then uniform convergence on compact regions of $\mathbb{C}$ implies that the limit function $f$ is holomorphic, and it vanishes identically on $\mathbb{R}$, so $f$ must be identically zero. But then applying $\mathbb{C}$-independence, we have that all the $a_i=0$.

This points in the direction of a counterexample when the $y_i$ are unbounded---if we let the $|y_i|$ tend to $\infty$ rapidly this argument fails dramatically.