Just two Naive constructions i thought about and that do not give simple Ore extensions.

We start with an algebraically closed field $F$ of characteristic $0$ and consider the ring of polynomials $R_0=F[y]$. As a homomorphism $\sigma_0:R_0\rightarrow R_0$, we use the evaluation at $0$, i.e. $\sigma_0(\sum_{i=0}^n a_iy^i)=a_0$. The $\sigma_0$-derivation is then given by $\delta_0(\sum_{i=0}^n a_iy^i)=\sum_{i=1}^n a_iy^{i-1}$.

The Ore extension $R_0[x,\sigma_0,\delta_0]$ is not simple because $yx-1$ generates a non-trivial ideal. But we can do the following.

Define $R=R_0[(y-a)^{-1}\mid 0\not=a\in F]$ to be the localization of $R_0$ with respect to all polynomials $P\in R_0$ with $P(0)\not=0$. The homomorphism $\sigma_0$ and the derivation $\delta_0$ extend uniquely to this localization, i.e. we find a unique homomorphism $\sigma:R\rightarrow R$ such that $\sigma\vert_{R_0}=\sigma_0$, and a unique $\sigma$-derivation $\delta:R\rightarrow R$ such that $\delta\vert_{R_0}=\delta_0$.

Now the Ore extension $A=R[x,\sigma,\delta]$ is still not simple. Consider the set $I=\{b\in A\mid \exists k\in\mathbb{N}:by^k=0\}$. This set is clearly a left ideal, it does not contain $1$ and it contains $yx-1$. It is also immediate that $I\cdot R=I$. In order to show that $I$ is a non-trivial ideal, we only have to show that $Ix\subset I$. Suppose $b\in I$, so $by^k=0$ for some $k$. Then we see that $(bx)y^{k+1}=by^k=0$ so $bx\in I$.