Let $X$ be a semimartingale under probability measure $P$ with representation $X = X_0 + M + A$ where $M$ is a martingale and $A$ is a finite variation process. Suppose $X = X_0 + N + B$ under $Q$, where $Q$ is another probability measure that is equivalent to $P$, $N$ is a $Q$-martingale, and $B$ is a finite variation process. Girsanov's theorem tells us how to relate $M$ and $N$, and $A$ and $B$, respectively.
Question: If we know that $A$ is actually absolutely continuous with respect to Lebesgue measure, does Girsanov tell us that $B$ must be so too?
The answer is no in general. You are assuming that $A\ll \lambda$, where $\lambda$ is the Lebesgue measure on the time axis. If the Lebesgue measure also dominates $\langle M,M\rangle$, so that $$\langle M,M\rangle\ll \lambda,\tag{1}\label{1}$$ then yes.
The following can be helpful. If there is an equivalent local martingale measure (ELMM) for $X$ then necessarily $A\ll \langle M,M\rangle$. If ELMM exists then necessarily $B \ll \langle M,M\rangle$. If additionally we know that \eqref{1} holds then you get what you wanted.