Let us consider a probability space $(\Omega,\mathscr{F},\mathbb{P})$ endowed with a filtration $\mathbb{F}$, and let $N$ and $M$ be two independent Poisson process with different parameter $\lambda$ and $\gamma$. We define the process $A$ as $A_t:=N_tM_t$. Then by Itô's Lemma for jump diffusions: \begin{align} \Delta A_t &=N_{t^-}\Delta M_t+M_{t^-}\Delta N_t+\Delta[N,M]_t\\[3pt] &=N_{t^-}\Delta M_t+M_{t^-}\Delta N_t+\Delta N_t\Delta M_t\\ \end{align}
Under which conditions can be assume that $N$ and $M$ do not jump simultaneously, that is $\Delta N_t\Delta M_t=0$? Intuitively, if $N$ and $M$ are independent, then the probability that some jump time $\tau$ from process $N$ coincides with another jump time $\tau^\prime$ from $M$ is vanishingly small $-$ but I don't think that is enough to state $\Delta N_t\Delta M_t=0$, a rigorous proof would be needed. However given there are infinitely many jumps over $[0,\infty)$ from both processes, it could be that over a sufficiently large time two jump times might coincide.
Let now $J^N$ be another jump process defined as $J^N_t:=1_{N_t\geq1}$ that is $J^N$ jumps to 1 whenever $N$ jumps for the first time. We similarly define $J^M$. These two processes jump at most once over $[0,\infty)$ $-$ whereas $N$ and $M$ can jump infinitely many times. Similarly, under which conditions can be assume that $\Delta J^N_t\Delta J^M_t=0$?
Here supply a direct proof of the fact $\Delta N_t\Delta M_t =0 $.
Suppose $S,T$ are two independent random variables with distribution funtions $F_S,F_T $ respectively, then \begin{align*} \mathsf{P}(S=T) &= \int_{\mathbb{R}}\int_{\mathbb{R}} 1_{\{s=t\}}(s,t)\,\mathrm{d}F_S(s)\,\mathrm{d}F_T(t) =\int_{\mathbb{R}}\Delta F_S(t)\,\mathrm{d}F_T(t)\\ &=\sum_t \Delta F_S(t)\Delta F_T(t). \end{align*} In particule, if one of $F_S,F_T$ is a continuous distribution, then $\mathsf{P}(S=T)=0$.
Now suppose that the successive jump times of independent Poisson Processes $M,N$ are $\{S_m,m\ge 1\}, \{T_n, n\ge 1\} $ respectively. The distributions of $\{S_m,m\ge 1\}, \{T_n, n\ge 1\} $ are $\Gamma$-distributions, which have continuous DFs. Hence, \begin{equation*} \mathsf{P}(S_m=T_n)=0,\qquad \forall m,n\in \mathbb{N}_+. \tag{1} \end{equation*} Meanwhile \begin{equation*} \{(\omega,t):\Delta M_t(\omega)\Delta N_t(\omega)\neq 0 \} =\bigcup_{m,n}\{(\omega,t):[\hskip-1.5pt[S_m]\hskip-1.5pt]=[\hskip-1.5pt[T_n]\hskip-1.5pt]\} \end{equation*} (1) means $\{(\omega,t):[\hskip-1.5pt[S_m]\hskip-1.5pt] = [\hskip-1.5pt[T_n]\hskip-1.5pt]\} $ is evanescent, i.e., \begin{equation*} \mathsf{P}(\{\omega: \exists t>0 \text{ with } t=S_m(\omega)=T_n(\omega)\})=0. \end{equation*}
Furthermore, \begin{equation*} \Delta M\Delta N =0\;(\mathsf{P}(\{\omega: \Delta M_t(\omega)\Delta N_t(\omega)=0, \forall t>0 \})=1) . \end{equation*}
Add in Proof: About the "evanescent" and $[\hskip-1.5pt[\; \cdot \;]\hskip-1.5pt]$, please refer to J. Jacod, and A. N. Shiryayev, Limit Theory for Stochastic Processes, 2ed. Springer, 2003. p.3("evanescent") and p.6($[\hskip-1.5pt[\; \cdot\; ]\hskip-1.5pt]$).