Not necessarily. For example, consider
$$ A = \pmatrix{\cos(\theta) & -\sin(\theta)\cr \sin(\theta) & \cos(\theta)}$$
with eigenvalues $e^{\pm i \theta}$ having positive real part if $-\pi/2 < \theta < \pi/2$.

Since $A$ is orthogonal, its SVD has $U = A$, $\Sigma = V^T = I$.
Then
$$A' = A \pmatrix{\sigma_1' & 0 \cr 0 & \sigma_2'}$$
has trace $(\sigma_1' + \sigma_2') \cos(\theta) > 0$ and determinant
$\sigma_1' \sigma_2' > 0$, and therefore its eigenvalues have positive
real part.

EDIT:
On the other hand, it is not always impossible. There are $3 \times 3$ examples where the real part
of a complex-conjugate pair of eigenvalues changes sign depending on the $\sigma'_j$. For example, take
$$ \eqalign{U &= \pmatrix{\sqrt{18-6\sqrt{3}}/6 & 0 & -\sqrt{18+6\sqrt{3}}/6 \cr
-\sqrt{9+3\sqrt{3}}/6 & \sqrt{2}/2 & -\sqrt{9-3\sqrt{3}}/6\cr
\sqrt{9+3\sqrt{3}}/6 & \sqrt{2}/2 & \sqrt{9-3\sqrt{3}}/6\cr}\cr
V^T &= \pmatrix{-\sqrt{9-3\sqrt{3}}/6 & \sqrt{9-3\sqrt{3}}/6 & \sqrt{18+6\sqrt{3}}/6\cr \sqrt{2}/2 & \sqrt{2}/2 & 0\cr
-\sqrt{9+3\sqrt{3}}/6 & \sqrt{9+3\sqrt{3}}/6 & -\sqrt{18-6\sqrt{3}}/6\cr}}$$
which come from the SVD of $$\pmatrix{0 & 0 & 1\cr 1 & 0 & -1\cr 0 & 1 & 1\cr}$$
For some $(\sigma'_1, \sigma'_2, \sigma'_3)$, such as $(1,1,0.3)$, the eigenvalues all have positive real part; for others, such as $(1,1,0.4)$,
the real eigenvalue is positive but the complex conjugate pair have negative real part.