To get some feeling for the problem, we make the following simplifying
assumptions:

$B$ is independent of $x$, say $B \in \mathcal S(\overline{\mathbb R}_+)$.

$A$ admits a (holomorphic) functional calculus.

There is a (holomorphic) function $h\colon \sigma(A)\to\overline{\mathbb H}_-$ such that $\bigl(i\tau - \lambda\,\hat{B}(\tau)\bigr)\bigr|_{\,\tau=h(\lambda)} =0$ for $\lambda\in\sigma(A)$, where $\hat{B}(\tau)=\int_0^\infty e^{-it\tau}B(t)\,dt$ is the Fourier transform of $B$.

Then it is readily seen that
$$
u(t) = e^{ith(A)} u_0
$$
is a solution of the original problem. Notice that, in general, *one can only solve backwards in time*.

There are several difficult points with this approach, the two obvious ones being:

The range of $i\tau\,/\,\hat{B}(\tau)$ might miss parts of the spectrum of $A$. This puts restrictions on the initial value $u_0$.

There might be several choices for the function $h$. This leads
to non-uniqueness of the solution $u$.

Here is an **example**: Take $B(t) = H(t)\,e^{-t}$. Then $\hat{B}(\tau)= \frac{1}{1+i\tau}$. Choose $A\in \operatorname{Op}S^2$ with principal symbol $|\xi|^2$ and $\sigma(A)\subset\{\lambda\in{\mathbb C}\mid \Re \lambda>0\}$. Then $h(\lambda) = \frac{i}2\left(1-\sqrt{1+4\lambda}\right)$. It follows that $h(A)\in \operatorname{Op}S^1$ with principal symbol $-i\,|\xi|$. This implies that the operators $e^{ith(A)}$ for $t<0$ belong to $\operatorname{Op}S^{-\infty}$, i.e., they are regularizing, and *there is no propagation of singularities*.

ADDED: I'm not aware of any holomorphic functional calculus for pseudodifferential operators - with $m>0$ and not necessarily having $A$ be normal (i.e., $AA^*=A^*A$ as unbounded operators in $L^2$) - which does what is required here.

Indeed, $\hat B$ extends to a $C^\infty$ function in $\overline{\mathbb H}_-$ which is holomorphic in $\mathbb H_-$, with
$$
\hat B(\tau) \sim \sum_{k\geq 0} \frac{B^{(k)}(+0)}{(i\tau)^{k+1}} \quad
\text{as $\tau\to \infty$ in $\overline{\mathbb H}_-$}
$$
(and this asymptotic expansion can be differentiated any number of times). Therefore, assuming $B(+0)\neq0$, one has
$$
\frac{i\tau}{\hat{B}(\tau)} = -\,\frac{\tau^2}{B(+0)} + O(\tau) \quad \text{as
$\tau\to\infty$ in $\overline{\mathbb H}_-$.}
$$
So, one would expect $h(A)$ to belong to $\operatorname{Op}S^{m/2}$ and to have principal symbol
$$
-i\sqrt{B(+0)} \ a_m^{1/2}(x,\xi),
$$
where $a_m(x,\xi)$ denotes the principal symbol of $A$.

Of course, *$h(\lambda)$ need not be an algebraic function of $\lambda$* - as was the case in the example above - and then in order to prove such a result one cannot directly appeal to known facts about complex powers of (hypoelliptic) pseudodifferential operators.

Still, as
$$
h(\lambda) \sim \sum_{l\geq0} c_l\lambda^{1/2-l} \quad\text{as $\lambda\to\infty$}
$$
in a suitable sector depending on the choice of $B$ (again, this asymptotic expansion can be differentiated any number of times), with $c_0 = -i\sqrt{B(+0)}$ as just seen and $c_l$ for $l\geq0$ computable in terms of $B(+0),\dots, B^{(l)}(+0)$, there is a good chance that such a result holds. See also this previous post.

ADDED: Suppose that $e^{at}B\in \mathcal S(\overline{\mathbb R}_+)$ holds for some $a>0$. Then $\hat{B}$ extends to a $C^\infty$ function on $ia+\overline{\mathbb H}_-=\{\tau\in \mathbb B\mid\Im\tau\leq a\}$ which is holomorphic on $ia+\mathbb H_-$ (and still has an asymptotic expansion as $\tau\to\infty$, as above).

Here is another **example** using this observation: Let $B(t)=-\,H(t)\,e^{-t}$ and $A\in\operatorname{Op}S^2$ with principal symbol $|\xi|^2$. Suppose that $\sigma(A)\subset \{\lambda\in{\mathbb C}\mid (\Im \lambda)^2<\Re\lambda\}$. Then $i\tau/\hat{B}(\tau)=\tau^2-i\tau$ and there are the two choices $h_\pm(\lambda)= \frac12\left(i\pm \sqrt{4\lambda-1}\right)$ for $h$. Furthermore, $h_\pm(A) \in \operatorname{Op}S^1$ with principal symbol $\pm|\xi|$. To specify a unique solution $u$ requires a second initial condition $u_t(0)=u_1$. Contrary to the situation considered before, the solution
$$
u(t) = e^{-t/2}\left(\cos\left(t\sqrt{A-1/4}\right)u_0 +
\frac{\sin\left(t\sqrt{A-1/4}\right)}{\sqrt{A-1/4}}\left(\frac{u_0}2+u_1\right)\right)
$$
is defined for all $t\in\mathbb R$. Moreover, now *one has propagation of singularities*.