If all $m_{i,\lambda}\geq 0$, the answers to all these questions are "yes".
The key observation is that
$1/(s-\lambda)$ has positive real part when $s$ is in the right half-plane, while $\lambda$
is in the left half-plane. Therefore even if your series converges at ONE point $s$,
then $\Re f_j(s)$ is a series with positive terms, from which all your statements immediately follow. Same if all $m_{i,\lambda}\leq 0$.

EDIT. In general, without positivity assumption, this is not true. The following example was given by user fedja.

For $n\geq 2$, let $p_n$ be rational functions with single (multiple)
pole at $-n$,
$p_n(z)\to \phi$ uniformly in the right half-plane.
Here $\phi$ is an arbitrary bounded
analytic function
in the half-plane $\Re z>-1$.

Such functions $p_n$
are polynomials of $1/(z+n)$ and they exist by Runge's theorem
(in fact one does not need Runge's theorem for approximating on a disk).
For each $n$, let $Q_n$ be polynomials, partial sums of $e^z$, of sufficiently
high degree $d_n$. The degree is such that the zeros of $Q_n\circ p_n$
are all in the disc $|z+n|<1/2$. This is possible because $Q_n(z)\to e^z$
as $d_n\to\infty$ uniformly on compact subsets of the plane,
and $e^z$ has no zeros. (Zeros of $Q_n$ tend to $\infty$ but $p_n$ is close to $\infty$
only near the point $-n$.)

Consider now the rational functions $g_n=Q_n\circ p_n$. They converge
to the function $e^\phi$ uniformly in the right half-plane.

Zeros and poles of $g_n$ are in the left half-plane, and they
tend to $\infty$ as $n\to\infty$. We take the set of all these zeros
and poles as our sequence $\Lambda$.

Finally, $$F_n(z):=g_n'(z)/g_n(z)=
\sum_{\lambda\in\Lambda}\frac{m_{\lambda,n}}{z-\lambda},$$
where the sum is finite, and $m_{\lambda,n}$ are integers. This sequence tends
to $\phi'$ uniformly in the right half-plane. But $\phi'$
was more or less arbitrary, so you can arrange it to have any singularities
in the left half-plane.
Now convert the sequence $F_n$ into a series, if you wish.