For every $t\geqslant0$, let $X_t=\sum\limits_{i=1}^{N_t}J_i\cdot Y^i_{t-T_i}$. A simple example is when the sequence $(J_i)_{i\geqslant i}$ is i.i.d. Bernoulli $\pm1$ centered and, as suggested in the question, $(Y^i)_{i\geqslant i}$ are i.i.d. standard Brownian motions. Then the process $(X_t)_{t\geqslant0}$ is distributed like $(Z_t)_{t\geqslant0}$ defined by $ Z_t=\sum_{i=1}^{N_t}J_i+\int_0^t\sqrt{N_t}\mathrm dY_s, $ where $Y=(Y_t)_{t\geqslant0}$ is a standard Brownian motion. In words, conditionally on $Z_t=z$ and $N_t=n$, when $s\to0^+$ and neglecting $o(s)$ terms, $Z_{t+s}=z+\varepsilon H+\sqrt{ns}G$ and $N_{t+s}=n+H$ for some independent $(H,\varepsilon,G)$, where $H=1$ and $H=0$ with probability $s$ and $1-s$ respectively, $\varepsilon=+1$ and $\varepsilon=-1$ with probability $\frac12$ and $\frac12$ respectively, and $G$ is a standard centered normal random variable.
Thus, $(Z_t,N_t)_{t\geqslant0}$ is a Markov process on the state space $\mathbb R\times\mathbb N_0$.
For every function $u$, let $Du:z\mapsto \frac12(u(z+1)+u(z-1)-2u(z))$ denote the discrete Laplacian of $u$. The dynamics described above yields the differential equation \frac{\mathrm d}{\mathrm dt}\mathrm E(u(Z_t))=\mathrm E(Du(Z_t))+\frac12\mathrm E(N_tu''(Z_t)). In particular, $ \left.\frac{\mathrm d}{\mathrm ds}\mathrm E(Z_{t+s}^2\mid Z_t,N_t)\right|_{s=0}=1+N_t. $ The proof is complete if $N_t$ is not measurable for $Z_t$. But the conditional distributions of $Z_t$ conditionally on $N_t=n$, for every $n\geqslant1$, are all absolutely continuous with respect to each other (the case $n=0$ being different since, conditionally on $N_t=0$, $Z_t=0$ almost surely), hence one can reconstruct $N_t$ from $(Z_s)_{0\leqslant s\leqslant t}$ (count the jumps) but not from $Z_t$ alone.
This proves that $\mathrm E(Z_{t+s}^2\mid Z_t,N_t=n)$ does depend on $n$, at least when $s\to0^+$, hence that $(Z_t)_{t\geqslant0}$ is not a Markov process with respect to its own filtration.