For an autonomous system:
$$x'= Ax$$
This solution is asymptotically stable as $t \to \infty$ if and only if for all eigenvalues $λ$ of $A$, $\Re(λ) < 0$.
In that case, would the system:
$$x'= Ax + Bx$$
keep stable, if all the eigenvalues of B are pure imaginary?
No, in general there's no reason to expect the eigenvalues of the sum of two matrices to be equal to or even controlled by the sum of the eigenvalues of the two matrices (except for the fact that traces add). Explicitly, take
$$A = \left[ \begin{array}{cc} -1 & t \\ 0 & -1\end{array} \right], B = \left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right].$$
where $t$ is an indeterminate. The eigenvalues of $A$ are $-1, -1$ and the eigenvalues of $B$ are $i, -i$, but
$$A + B = \left[ \begin{array}{cc} -1 & t + 1 \\ -1 & -1 \end{array} \right]$$
has characteristic polynomial $\lambda^2 + 2 \lambda + (t + 2)$ and hence eigenvalues
$$\lambda = \frac{-4 \pm \sqrt{4 - 4(t + 2)}}{2}.$$
By taking $t \to - \infty$ we can arrange for the larger eigenvalue to get not only positive but arbitrarily large.
Things are better if you assume either that $A$ and $B$ commute or that $A$ and $B$ are self-adjoint.