No you cannot use biot savart law.
Biot savart law is derived using the equation:
$ \int B \cdot dl = \mu_0 I$
$\nabla × \vec{B} = \mu_0 \vec{J}$
taking the divergence of this gives
$\nabla \cdot J = 0$
the current density of a point charge doesn't follow this.
You are probably only familiar with the line integral version.The biot savart law is actually a volume integral, using current density instead.
The issue isn't that a path isnt defined as we can easily use a volume integral. The issue is that we are deriving under the assumption that the divergence of current density is zero.
Naively plugging the current density of a point charge into biot savart law looks like the following (through the magnetic vector potential):
$\nabla × A = B$
$\nabla × (\nabla × A) = \mu_0 J$
$\nabla(\nabla \cdot A) - \nabla^2 A = \mu_0 J$
Set $\nabla \cdot A = 0 $ due to field invariance
$\nabla^2 A = -\mu_0 J$
$ A(r',t) =\iiint \frac{\mu_0}{4\pi} \frac{J(r',t)}{|r-r'|} d^3r'$
for a point charge
$J(r',t) = Q r_{s}'(t)\delta^3(r'-r_{s}(t))$
where $ r_{s}(t) $is the position vector of the charge
A =$\iiint \frac{Q\mu_0}{4\pi} \frac{ r_{s}'(t)\delta^3(r'-r_{s}(t))}{|r-r'|} d^3r'$
= $\frac{Q\mu_0}{4\pi} \frac{r_{s}'(t)}{|r-r_{s}(t)|}$
$B = \nabla × (\frac{Q\mu_0}{4\pi} \frac{r_{s}'(t)}{|r-r_{s}(t)|})$
Changing coordinates such that$ |r-r_{s}(t)|$ is just radial distance R from the charge,
$ \frac{Q\mu_0}{4\pi} \nabla × ( \frac{r_{s}'(t)}{R})$
=$ \frac{Q\mu_0}{4\pi} (\nabla(\frac{1}{R}) × r_{s}'(t) +\frac{1}{R}\nabla × r_{s}'(t) ) $
=$ \frac{Q\mu_0}{4\pi} \nabla(\frac{1}{R}) ×r_{s}'(t) $
=$ -\frac{Q\mu_0}{4\pi} \frac{\hat r ×r_{s}'(t)}{R^2} $
=$ -\frac{Q\mu_0}{4\pi} \frac{\hat r ×v}{R^2} $
=$ \frac{Q\mu_0}{4\pi} \frac{v×\hat r}{R^2} $
This is the "quasi static" formula.
The real and correct formula, is derived from the lienard wichert potentials, using amperes law with maxwells displacement current.
