A former professor of me suggested the following argumentation:
First, we rewrite the polynomial$$\begin{align}S(x) & = \prod_{j=1}^n (x + \lambda_{j}) +\sum_{i=1}^n k_i \cdot \prod_{j=1, j\neq i}^n (x + \lambda_{j}) \\&= \sum_{i=1}^n \left( k_i \cdot \hspace{-2mm} \prod_{j=1, j\neq i}^n (x + \lambda_{j})+ \frac{1}{n} \prod_{j=1}^n x + \lambda_{j} \right) \notag \\&=\sum_{i=1}^n \left( \prod_{j=1, j\neq i}^n (x + \lambda_{j}) \right)\cdot\left( \frac{x + \lambda_{i}}{n} + k_i \right) \\&=\sum_{i=1}^n \frac{1}{n}\left( \prod_{j=1, j\neq i}^n (x + \lambda_{j}) \right)\cdot\left( x + \lambda_{i} + n\cdot k_i \right).\end{align}$$
Now we let $x = \mu e^{\alpha i}$ with $\mu, \alpha \geq 0$, i.e. $x$ is a complex number in $\mathcal Q_1$, the first quadrant of the complex plain.
Clearly, we have $x + \lambda_j \in \mathcal Q_1$.Further, we remind a) that the angle-arguments of complex numbers add up when multiplying these complex numbers and b) that $\mathrm{arc}(\frac{1}{n} e^{i \alpha}) = \frac{\alpha}{n}$.
Now we see that$$\frac{1}{n}\left( \prod_{j=1, j\neq i}^n (x + \lambda_{j}) \right)\cdot\left( x + \lambda_{i} + n\cdot k_i \right) \in \mathcal Q_1$$holds, because it is the average of $n$ complex numbers from $\mathcal Q_1$.
Finally, $S(x)$ is just the sum of $n$ of such "averages" and thus also in $\mathcal Q_1$.Because $k_i >0$ is a strict inequality, the value $S(x)$ can not be zero.This proves $S(\mathcal Q_1) \neq 0$. For the fourth quadrant an analogous argument holds.
