Expanding the comment by Alexandre Eremenko.

Let us rewrite the equation in the form $$ \underbrace{\prod_{j=1}^n (x-x_j)}_{f(x)} + \underbrace{i \sum_{j=1}^n \lambda_j \prod_{s\ne j} (x-x_s)}_{g(x)} = 0. $$

Choose $r = \frac12 \min_{j\ne s} |x_j - x_s|$ and consider $B_r(x_k)$. Obviously $f(x)=0$ has a root inside $B_r(x_k)$. Moreover $\min_{x\in \partial B_r(x_k)} |f(x)| > 0$. Let us estimate $|g(x)|$ in terms of $|f(x)|$ for $x\in \partial B_r(x)$.

$$ |g(x)| \le \sum_{j=1}^n \frac{|\lambda_j|}{|x-x_j|} |f(x)| \le \frac{\sum_{j=1}^n |\lambda_j|}{r} |f(x)| < |f(x)| $$

provided $\frac1r \sum_{j=1}^n |\lambda_j| < 1$, which holds if $\lambda_j$ are sufficiently small for all $j$. Then it remains to apply Rouché's theorem.

Expanding the comment by Alexandre Eremenko (and assuming $\lambda_j$ small for $j=k$ only).

Let us rewrite the equation in the form $$ \underbrace{\prod_{j=1}^n (x-x_j) + i \sum_{j\in\{1,\ldots,n\} \setminus\{k\}} \lambda_j \prod_{s\in\{1,\ldots,n\} \setminus \{j\}} (x-x_s)}_{f(x)} + \underbrace{i \lambda_k \prod_{s\in\{1,\ldots,n\} \setminus \{k\}} (x-x_s)}_{g(x)} = 0. $$

Choose $r>0$ sufficiently small so that $f(x) \ne 0$ for $x\in\partial B_r(x_k)$. Then $\min_{x\in \partial B_r(x_k)} |f(x)| = m > 0$. On the other hand if $\lambda_k$ is sufficiently small then $|g(x)| < m$ for all $x \in \partial B_r(x_k)$. Since $f(x_k)=0$ it remains to apply Rouché's theorem.