Is the image of a tensor product equal to the tensor product of the images?

No, because the map $im(\phi)\otimes_S im(\psi) \to A'\otimes_S B'$ may not be injective.

E.g. consider the case $S = \mathbb Z$, $A = A' = B =\mathbb Z/2\mathbb Z,$ $B' = \mathbb Q/\mathbb Z$, $\phi = id_A$, and $\psi: \mathbb Z/2\mathbb Z \hookrightarrow \mathbb Q/\mathbb Z$ is the unique injection.

Then the map $A \otimes_S B \to im(\phi) \otimes_S im(\psi)$ is an isomorphism, while the map $\phi \otimes_S \psi$ is the zero map, since $A'\otimes_S B' = 0$.

More generally, if you restrict to the case when $\phi$ is the identity, and $\psi$ is injective, you are asking whether the injection $\psi: B \hookrightarrow B'$ induces an injection $A\otimes_S B \hookrightarrow A \otimes_S B'$. The answer is no in general because tensoring is not left exact. (The preceding example is one illustration of this.)