A metric of the form
$$ds^2 = - \left( \frac{c dt}{A(t)}\right)^2 + ds_3^2 ~~~~(*)$$
is a flat metric, which you can see by defining a new variable
$$ t' = \int^{t'} \frac{c dt}{A(t)},$$
so that
$$ds^2 = - (dt')^2 + ds_3^2. $$
Therefore there is no choice of ##A(t)## such that this metric is equivalent to the FLRW metric.
$$ ds^2 = - c^2 dt^2 + a(t)^2 ds_3^2 .$$
If we were dealing with the open or closed versions of ##ds_3^2## in (*), we'd find that the curvature of the 4-metric was independent of ##A(t)## and was just given by the spatial curvature.
An equivalent form to the FLRW metric is
$$ ds^2 = a(\tau)^2 \left( - d\tau^2 + ds_3^2 \right),$$
which can be obtained by defining the so-called conformal time
$$ \tau = \int^{\tau} \frac{c dt}{a(t)}.$$
The problem with the form that you postulate is that the coordinate transformation you have to make to remove the ##a(t)## from the spatial part involves mixing the time variable with the radial variable:
$$ r' = a(t) r.$$
However, now ##dr'## involves a term with ##dt## and you'll inevitably find cross terms in the transformed metric of the form ##dt' dr'## that don't vanish (these terms preserve the curvature). There doesn't appear to be an appropriate choice of ##t'## such that you end up with the simple expression that you're hoping for, even if we were willing to let the function ##A=A(r)##.
