Yes it is possible already in classical physics.
In classical physics, spacetime is a 4-dimensional affine space $\mathbb{A}^4$ equipped with a surjective affine map $T: \mathbb{A}^4 \to \mathbb{R}$ defined up to additive constants corresponding to absolute time. The 3-planes $\Sigma_t := \{p \in \mathbb{A}^4 \:|\: T(e)=t\}$ are the absolute spaces at time $t$. They are equipped with a metric structure (a positive scalar product in the 3-space of translations of each $\Sigma_t$) smoothly depending on the global affine structure.
In this picture, inertial reference frames are defined by a subclass of 4-dimensional Cartesian coordinates such that, in particular, one coordinate is $T$ itself (*) and the remaining three coordinates are tangent to the planes $\Sigma_t$ and form a 3-dimensional Cartesian orthonormal system of coordinates in each $\Sigma_t$.
As a consequence, the relations between a pair of these Cartesian coordinates is a generic Galileian transformation
$$x'^0= x^0+c$$
$$x'^k = c^k+ tv^k + \sum_{h=1}^3{R^k}_h x^h\quad k=1,2,3$$
where $c, c^k, v^k \in \mathbb{R}$ and $[{R^h}_k]_{h,k=1,2,3} \in O(3)$.
Then, it is possible to define a 4-dimensional affine connection $\nabla$ in $\mathbb{A}^4$ just by declaring the connection coefficients of $\nabla$ vanish in these (inertial) reference frames. Evidently a curve $\gamma$, given by $$t \mapsto (t, x^1(t),x^2(t), x^3(t))$$ in inertial coordinates, is a geodesic with respect to $\nabla$ if and only if it is an inertial motion
$$x^k(t) = u^k t + c^k\quad k=1,2,3\:.$$
When using other reference frames associated with coordinates $y^\alpha$, the equation reads in full generality but keeping the parametrization in terms of absolute time,
$$\frac{d^2y^\mu}{dt^2} = -\Gamma^\mu_{\alpha \beta} \frac{dy^\alpha}{dt} \frac{dy^\beta}{dt}\:.\tag{1}$$
Above, directly from the transformation law of the connection coefficients $$\Gamma^\mu_{\alpha \beta} = \sum_{\sigma,\delta=0}^3\frac{\partial^2 y^\mu }{\partial x^\sigma \partial x^\delta} \frac{\partial x^\sigma}{\partial y^\alpha} \frac{\partial x^\delta}{\partial y^\beta} \tag{2}$$
with $x^\mu$ being Cartesian coordinates of any inertial reference frame.
A generic reference frame of the classical physics with coordinates $y^\alpha$ is related to an inertial reference frame with coordinates $x^\beta$ through the general relation
$$y^0=x^0 + c \quad (=t+c')\:,$$
$$y^k = c^k(x^0) + \sum_{h=1}^n {R^k}_h(x^0) x^k \quad k=1,2,3\:.$$
Above, the maps $\mathbb{R} \ni t\mapsto [{R^k}_h(x^0)] \in O(3)$ and
$\mathbb{R} \ni t\mapsto c^k(x^0) \in \mathbb{R}$ are smooth.
If we use there relations to compute the coefficients $\Gamma^\mu_{\alpha \beta}$ according to (2), in particular defining
$$\omega^p := -\frac{1}{2} \sum_{k,q,s=1}^3 \epsilon_{pqs} {R^q}_k \frac{d{R^s}_k}{dt}\quad p=1,2,3$$
Eq.(1) turns out to become, where we also multiply with the mass $m$ of the point,
$$m\frac{d^2 \vec{y}}{d t^2} =- m \vec{a}(t)- 2 m\vec{\omega} \wedge \vec{v} - m\vec{\omega}\wedge (\vec{\omega} \wedge \vec{y})-m \dot{\vec{\omega}} \wedge \vec{y}$$
where $v^k = \frac{dy^k}{dt}$, and $\vec{a}(t)$ is the acceleration of the origin of the coordinates $y^\alpha$ computed in the inertial coordinates $x^\mu$ and finally expressed in the basis of the reference frame $y^\alpha$.
It is now clear that, within this formalism, the connection coefficients account for the inertial forces.
(*) This means that the vector $e_0$ defining that coordinate satisfies $\langle e_0, dT \rangle =1$.
