Let $0 < \lambda_1 \leq \ldots \leq \lambda_n $ and $k_1, \ldots, k_n> 0$.
Let further $$\begin{align}P(x)&:=\prod_{i=1}^n (x+\lambda_i) = (x+\lambda_1)\cdot \ldots \cdot (x+\lambda_n) \\q_i(x) &:= \frac{P(x)}{(x+\lambda_i)}\\S(x) &:= P(x) + \sum_{i=1}^n k_i q_i(x)\end{align}$$be polynomials in the complex valued unknown $x$.
E.g. for $n=2$ we have
$$S(x) = (x + \lambda_1)(x + \lambda_2) + k_1(x+\lambda_2) + k_2(x+\lambda_1).$$
I suspect that S(x) has no zeros with positive real part. I tried it numerically 100 times for $n=5$ with random numbers for $x_i, \lambda_i$ and always got 5 negative real roots. (So, I guess that actually there are no non-real roots, but that is not so important.)
From Descartes rule of signs we know that $S$ has no positive real roots (because all coefficients of $S$ are positive). But how can we exclude complex roots with positive real parts (or as a whole)?
Update:
For the special case $\lambda_1 =\ldots = \lambda_n =:\lambda <0$ the assertion is almost trivial. In this case we have
$$S(x) = (x+ \lambda)^n + \sum_{i=0}^n k_i (x + \lambda)^{n-1} = (x + \lambda)^{n-1} (x + \lambda + \sum_{i=0}^n k_i),$$i.e., we have $n-1$ zeros unchanged at $\lambda$ and one zero shifted by $\sum_{i=0}^n k_i$ to the left.