The two question are very much related; I hope this goes some way to help you. I have been suitably vague at points to hopefully get the ideas across but happy to try and answer any further concerns (to the best of my ability!).
1) Given a vector field $X$, the question arises: does this give rise to a family of curves? That is, can we find a family of curves whose tangent vectors are precisely $X$.
Let $X=\xi^{1}(x,y)\frac{\partial}{\partial x} + \xi^{2}(x,y)\frac{\partial}{\partial y}$ for some smooth functions $\xi^{i}(x,y)$ and define a curve on $\mathbb{R}^{2}$ by:
$$\begin{align}
\gamma:[a,b] &\longrightarrow \mathbb{R}^{2} \\
t &\longmapsto (x,y) = (\gamma^{1}(t), \gamma^{2}(t)).
\end{align}$$
The vector field along $\gamma$ is then
$$V^{\gamma} = \gamma_{*}\frac{d}{dt} = \frac{d\gamma^{1}}{dt}\frac{\partial}{\partial x}+ \frac{d\gamma^{2}}{dt}\frac{\partial}{\partial y} $$
where $\gamma_{*}$ denotes the push-forward or tangent map. Thus $X$ is a tangent vector to the curve if the components of $X$ and $V^{\gamma}$ coincide on $\gamma$:
$$\frac{d\gamma^{1}}{dt} = \xi^{1}(\gamma(t)), \quad \frac{d\gamma^{2}}{dt} = \xi^{2}(\gamma(t)).$$
From the theory of ordinary differential equations, solutions to this system always exist and are uniquely determined by initial conditions: $\gamma^{1}(0)=x_{0}$ and $\gamma^{2}(0)=y_{0}$. The curve is then said to start at the point $p=(x_{0},y_{0})$. With the map $\gamma$ starting at $p=\gamma(0)$ defined (the integral curves of the vector field $X$), we may then define
$$\begin{align}
\varphi_{t}:\mathbb{R}^{2} &\longrightarrow \mathbb{R}^{2} \\
p &\longmapsto \varphi_{t}(p) = \gamma(t)
\end{align}$$
which can be shown to be a local one-parameter family of diffeomorphisms with the group structure you are asking for.
If you try this with your vector field $aX=ax\frac{\partial}{\partial y} - ay\frac{\partial}{\partial x}$, you get precisely the map $\rho:t\mapsto (\cos(at),\sin(at))$ provided the initial conditions $\rho^{1}(0)=1, \rho^{2}(0)=0$ are used. This then leads to a one-parameter family of diffeomorphisms $\varphi_{t}$ (a one-parameter group) as outlined above.
2) The second question is now the converse of the first: given a one-parameter family of diffeomorphisms, how do you find its associated vector field?
In your example, you are working purely on $\mathbb{R}$, so let us stay there (hopefully extensions to higher dimensions should be obvious).
We have a one-parameter family of diffeomorphisms:
$$\begin{align}
\varphi_{t}: \mathbb{R} &\longrightarrow \mathbb{R} \\
x &\longmapsto y = \varphi_{t}(x)
\end{align}$$
This induces a curve $\varphi_{x}$ on $\mathbb{R}$:
$$\begin{align}
\varphi_{x}: [0,a] &\longrightarrow \mathbb{R} \\
t &\longmapsto y = \varphi_{x}(t) = \varphi_{t}(x)
\end{align}$$
which starts at $x$ (i.e. $\varphi_{x}(0)=\varphi_{0}(x)=x$). Thus, we may define a curve starting at every $p\in\mathbb{R}$. Given the set of all such curves we may then define a tangent vector at each point in $\mathbb{R}$. The tangent vector to $\varphi_{x}$ is simply
$$X = \varphi_{x*}\frac{d}{dt} = \left.\frac{d\varphi_{x}}{dt}\frac{\partial}{\partial x}\right|_{t=0}$$
Notice that the result is restricted to $t=0$ as this is where the curve is defined to start.
Using this method with $\varphi_{x}(t)=x+t$ yields the vector $X=\frac{\partial}{\partial x}$.
