Setup. Let $(\Omega,\mathfrak A,P)$ be a probability space supporting a Brownian motion $W$ with finite time horizon $T>0$. Denote by $\mathfrak F$ the agumented filtration induced by $W$. Denote by $S^2(P,\mathfrak F)$ the space of continuous adapted processes $Y$ satisfying \begin{equation*} E[\sup_{t\in[0,T]} |Y(t)|^2 ] < \infty, \end{equation*} $L^2(P,\mathfrak F)$ the space of adapted, square-integrable processes $Z$, i.e. \begin{equation*} \forall t\in [0,T]:~E[|Z(t)|^2 ] < \infty, \end{equation*} $I^2(W)$ the space of predictable processes $H$ such that \begin{equation*} E \big[\int_0^T |H(t)|^2~dt \big] < \infty. \end{equation*} Let be $a,b,c$ bounded predictable processes, and $F: [0,T] \times \mathbb R \times \mathbb R \times \Omega \rightarrow \mathbb R$ the map given by $F(t,y,z,.) = a + by +cz$.
Theorem. Under the above assumptions, for every $\xi \in L^2(P)$, there exists a unique pair $(Y,Z) \in S^2(P,\mathfrak F) \times I^2(W)$ such that, almost surely, \begin{equation} Y(t) = \xi + \int_t^T F(s,Y(s),Z(s)) ~ds - \int_t^T Z(s)~dW(s), \end{equation} for any $t\in [0,T]$.
Remark. There is a proof using Picard iteration. This proof works in more general scenarios. However, this argument is abstract. There is a better one for the here-considered affine case. For $a=b=c=0$, the statement follows readily using the martingale representation theorem. Else, use the transformations \begin{equation*} \tilde Y(t) = Y(t) \operatorname e^{\int_0^t b(s)~ds} + \int_0^t a(s) \operatorname e^{\int_0^s b(u)~du}~dt,\quad \tilde Z(t) = \operatorname e^{\int_0^t b(u)~du} Z(t), \end{equation*} \begin{equation*} \tilde \xi = \xi \operatorname e^{\int_0^T b(s)~ds} + \int_0^T a(s) \operatorname e^{\int_0^s b(u)~du}~dt,\quad \frac{dQ}{dP} \mid_{\mathfrak F(T)} = \mathcal E(c \bullet W)(T), \end{equation*} to translate the problem to a BSDE with zero generator $\tilde F = 0$. However, to use the result from that particular case, we have to assure that $\tilde \xi \in L^2(Q)$. In that case, we obtain a solution $(\tilde Y, \tilde Z)$ of the zero-generator BSDE under $Q$, and the corresponding pair $(Y,Z)$ solves the BSDE. But how can we deduce that $(Y,Z) \in S^2(P,\mathfrak F) \times I^2(W)$?
QUESTION: Can you answer the bold-marked questions?
IDEAS: $\frac{dQ}{dP} \mid_{\mathfrak F(T)}$ is not bounded; this makes the problem difficult i.m.h.o. Yet, we have $L^p(P)$, for any $p \ge 1$. Hence, if $\xi \in L^{2+\epsilon}(P)$ for some $\epsilon>0$, then $\xi \in L^2(Q)$.