I have an argument for (a) in the case $N \cap M = 1$, which implies $NM = N \times M$. Let $K$ be some minimal normal subgroup of $NM$. Then by considering the quotient $KN / N \unlhd NM / N \cong M$ either $KN \le N$ or $KN / N$ is minimal normal in $MN / N$, in the last case $[x,y] \in N$ for each $x \in K, y \in NM$ by assumption. Similar by looking at $NM / M$ we find $K \le M$ or $[x,y] \in M$ for each $x \in K, y \in NM$. As $[N, G] \le N$ and $[M, G] \le M$ by normality, in all cases we have $[x,y] \le M \cap N = 1$ for $x \in K, y \in NM$, which gives the claim.
If $N \cap M \ne 1$ maybe (b) can help in the sense that we can deduce that property $(*)$ then holds for $M \cap N$.
EDIT (5/11/19): I found a proof for the general case which works quite different. Let $K$ be some minimal normal subgroup in $NM$. Then as $N$ is normal we have that $[K, N] \le N \cap K$ is normal in $NM$, which implies $[K,N] = 1$ or $[K,N] = K$. Assume $[K,N] = K$. Let $L \le K$ be some minimal normal subgroup of $N$ in $[K,N]$ and consider $H = \langle L^g \mid g \in NM \rangle$, which is normal in $NM$. Hence $H = [K,N] = K$.
The groups $L^g$ are also minimal normal in $N$ by normality of $N$ and the fact that non-trivial homomorphic images are minimal normal in its image, hence $H$ is a direct product of some subset of these conjugages. By assumption the factors are all in the center of $N$, hence $K$ is in the center of $N$, i.e. $[K,N] = 1$ contradicting the assumption. Simlar we find $[K,M] = 1$. So $[K,MN] = [K,M][K,N] = 1$, i.e. $K$ is in the center of $NM$. $\square$