No. Trivial example: suppose $\mu$ is Lebesgue measure and $\nu(S)=\begin{cases}0&|S|=\omega\\\infty&|S|\geq\omega_1\end{cases}$. Then $\mu\times\nu=\nu$ (morally — the product is defined by the same formula), which is atomic.
However, if one requires $\mu$ and $\nu$ to be ($\sigma$-)finite measures, then the product is indeed atomless.
Let $\mu$ be atomless and $\nu$ arbitrary, and write $\lambda=\mu\times\nu$. We will show $\lambda$ lacks atoms, by transfinite induction on the rank over the generators.1 For this induction, it is useful to adopt a slightly stronger-appearing definition of atomlessness:
$\lambda$ is atomless if, for any $0<\delta<\epsilon<1$ and measurable $A$, there exists $B\subseteq A$ such that $\delta\lambda(A)\leq\lambda(B)\leq\epsilon\lambda(A)$.
Note that either definition of non-atomicity is preserved under finite unions/intersections/complements, so the only challenge is countable ones. Proving that the definitions are equivalent is a bit tricky, but luckily Wikipedia's page on atomic measures includes a sketch, so I'll defer to them and just jump straight to the induction.
For a base case, suppose $A=B\times C$ for some $B$, $C$. If $A$ is null, the claim is trivial, so suppose $0<\lambda(A)=\mu(B)\nu(C)$, i.e. $\mu(B)>0$. Since $\mu$ is atomless, there exists $D\subseteq B$ with relative measure at most between $\delta$ and $\epsilon$; taking Cartesian products preserves relative measure, so $D\times C$ witnesses that $A$ is not atomic. Thus no rank-0 sets are atomic.
If $\alpha$ is a limit ordinal, then any set of rank at most $\alpha$ in fact has lower rank, so it remains to show the successor ordinal case.
Suppose all rank-$\alpha$ sets are not atoms and $A$ is rank-$(\alpha+1)$ and has positive measure. Either $A=\bigcup_{j\in\omega}{A_j}$ or $A=\bigcap_{j\in\omega}{A_j}$, where each $A_j$ has rank at most $\alpha$. Since our definition of atomic is symmetric under taking complements, it suffices to assume the former. Moreover, we can disjointify the $\{A_j\}_j$: let $\tilde{A}_0=A_0$ and $\tilde{A}_J=\left(\bigcup_{j=0}^J{A_j}\right)\setminus\tilde{A}_{J-1}$ for $J>1$. This doesn't preserve rank, but it does preserve atomicity (see above). By hypothesis, each $A_j$ is not atomic; we now take them disjoint, so that $\lambda(A)=\sum_{j\in\omega}{\lambda(A_j)}$.
Fix $\delta$ and $\epsilon$: for each $j$, there exists $B_j\subseteq A_j$ such that $\delta\lambda(A_j)\leq\lambda(B_j)\leq\epsilon\lambda(A_j)$. By disjointness, taking a union in $j$ adds measures, so that $\delta\lambda(A)\leq\lambda\left(\bigcup_{j\in\omega}{B_j}\right)\leq\epsilon\lambda(A)$, as desired.
1 A quick reminder of rank for $\sigma$-algebras, where I adopt the slightly nonstandard convention that taking complements does not increase rank: let $\mathcal{F}$ be a $\sigma$-algebra generated by $\mathcal{G}$. Define the (transfinite) sequence \begin{align*}
\mathcal{F}_0&=\mathcal{G}\cup\{G^{\mathsf{c}}:G\in\mathcal{G}\} \\
\mathcal{F}_{\alpha+1}&=\mathcal{F}_{\alpha}\cup\left\{\bigcup_{j\in\omega}{F_j}:\forall j\,F_j\in\mathcal{F}_{\alpha}\right\}\cup\left\{\bigcap_{j\in\omega}{F_j}:\forall j\,F_j\in\mathcal{F}_{\alpha}\right\} \\
\mathcal{F}_{\alpha}&=\bigcup_{\beta<\alpha}{F_{\beta}}\quad\quad\alpha\textrm{ limit ordinal}
\end{align*} It is a fact that $\mathcal{F}=\mathcal{F}_{\omega_1}$ (exercise!).
If $F\in\mathcal{F}_{\alpha}$, then we say $F$ has rank (over $\mathcal{G}$) at most $\alpha$; the rank of $F$ is the minimal such $\alpha$.
Best Answer
This article answers your question even for the more general situation when the space is $\sigma$-finite and shows uniqueness of the desired decomposition.