Note that for fixed $t$, we have by the martingale representation theorem that
$$Y_t = \int_0^t I\bigl( s \in [0,t-1) \bigr) \, dB_s.$$
Note in particular that for $t\leq 1$ the indicator yields $0$ trivially. However, the integrand is not progressively measurable in the Brownian filtration, thus it is not an Ito process (and by uniqueness of the martingale represenation theorem we can also find no other representation - it is thus neither an Ito process nor a seminmartingale).
However, if we define the filtration $\tilde{\mathcal{F}}_t = \mathcal{B}(\mathbb{R}_{\geq0}) \otimes \mathcal{F}_{(t-1) \vee 0}$, then it is a martingale with respect to this filtration.