Yes, absolutely. Copied and pasted from the end of this answer:
Given any smooth function $F$, we can define a Hamiltonian vector field $X_F$, and a corresponding flow $\varphi_{X_F}$. Given some parameter value $t$, $\varphi_{X_F}(\Omega, t)$ constitutes a canonical transformation (i.e. a change of coordinates). We would then call $F$ the generator of the transformation $\varphi_{X_F}$.
Note the similarity to quantum mechanics: Given any self-adjoint operator $F$, we can define a family of unitary operators $\exp[iFt]$. Given some parameter value $t$, $\exp[iFt]$ constitutes a unitary transformation (i.e. a change of basis). We would then call $F$ the generator of the transformation $\exp[iFt]$.
That's the tl;dr version. In the remainder of my answer, I will try to explain what each of those things means with as little baggage as possible, and then I will explicitly demonstrate that the momentum observable is the generator of spatial translations.
In Hamiltonian mechanics, the space of possible states of the system is the phase space $\Omega$. A point $x\in \Omega$ can be labeled by a generalized position $q$ and a generalized momentum $p$, so $x=(q,p)$. Note that $q$ and $p$ are not functions - they are simply coordinates which label a point in phase space.
An observable $F$ is a smooth function $F:\Omega\times \mathbb R \rightarrow \mathbb R$ which maps a point in phase space (and possibly the time) to a real number. The two simplest examples of observables are the position observable $$\mathcal Q:\Omega \times \mathbb R \rightarrow \mathbb R$$ $$(q,p,t) \mapsto q$$ and the momentum observable $$\mathcal P:\Omega \times \mathbb R \rightarrow \mathbb R$$ $$(q,p,t) \mapsto p$$
Every smooth function $F$ corresponds to a Hamiltonian vector field $X_F$ given by
$$X_F = \frac{\partial F}{\partial p} \frac{\partial}{\partial q} - \frac{\partial F}{\partial q}\frac{\partial}{\partial p}$$
where I am using the differential geometric understanding of a vector as a directional derivative.
An integral curve $\Gamma:\mathbb R \rightarrow \Omega$ of a vector field $X$ can be obtained by solving the differential equation
$$\frac{d}{dt}\Gamma(t) = X\big|_{\Gamma(t)}$$ where the notation on the right means that the vector field $X$ is evaluated at the phase space point $\Gamma(t)$. Though the context is somewhat sophisticated, this is just the same as the differential equations idea of taking a vector field and finding all of the curves whose tangents are parallel to the vectors at every point.
Given a vector field $X$, we can define a corresponding flow $\varphi_X: \Omega\times \mathbb R \rightarrow \Omega$ as follows:
Start with a point $x_0\in \Omega$. Find the integral curve $\Gamma_{X,x_0}$ of $X$ which starts at $x_0$ (that is, $\Gamma_{X,x_0}(0)=x_0$). Then, we define
$$\varphi_X(x_0,t) = \Gamma_{X,x_0}(t)$$
Intuitively speaking, $\varphi_X$ takes points in the phase space and "pushes" them along the integral curves of $X$.
We are finally equipped to answer your question. Given any smooth function $F$, we can define a Hamiltonian vector field $X_F$, and a corresponding flow $\varphi_{X_F}$. Given some parameter value $t$, $\varphi_{X_F}(\Omega, t)$ constitutes a canonical transformation (i.e. a change of coordinates). We would then call $F$ the generator of the transformation $\varphi_{X_F}$.
Note the similarity to quantum mechanics: Given any self-adjoint operator $F$, we can define a family of unitary operators $\exp[iFt]$. Given some parameter value $t$, $\exp[iFt]$ constitutes a unitary transformation (i.e. a change of basis). We would then call $F$ the generator of the transformation $\exp[iFt]$.
Let's apply this to the momentum observable. $\mathcal P(q,p,t)=p$, so $$X_{\mathcal P} = \frac{\partial}{\partial q}$$ The integral curves can be found via $$\frac{d}{ds} \Gamma(s) = X_{\mathcal P}\big|_{\Gamma(s)}$$ which, in component form where $\Gamma(s) = (\Gamma^1(s),\Gamma^2(s))$, reads $$\frac{d\Gamma^1}{ds} \frac{\partial}{\partial q} = \frac{\partial}{\partial q} \implies \frac{d\Gamma^1}{ds} = 1$$ $$\frac{d\Gamma^2}{ds} \frac{\partial}{\partial p} = 0 \implies \frac{d\Gamma^2}{ds} = 0$$ which implies that, starting from the initial condition that $\Gamma(0)=(q_0,p_0)$, $$\Gamma^1(s) = q_0 + s$$ $$\Gamma^2(s) = p_0$$ From there, the corresponding flow is $$\varphi_{X_\mathcal P}(q,p,s) = (q+s,p)$$
We therefore see that this flow corresponds to spatial translations, and is generated by the momentum observable $\mathcal P$.