I will approach this from a more general perspective. Your specific question will be answered at the end of this post. Let $\lambda \in \mathbb{R}$ be a real parameter. The choice $\lambda = 0.069$ will correspond to your sequence. The general recurrence can be written as
$$
\begin{pmatrix} a_{n+1} \\ b_{n+1}
\end{pmatrix} = \begin{pmatrix}
1 & 0 \\ \lambda & 1
\end{pmatrix} \begin{pmatrix} a_{n+1} \\ b_n
\end{pmatrix} = \begin{pmatrix}
1 & 0 \\ \lambda & 1
\end{pmatrix} \begin{pmatrix}
1 & -\lambda \\ 0 & 1
\end{pmatrix}\begin{pmatrix} a_n \\ b_n
\end{pmatrix} = \begin{pmatrix}
1 & -\lambda \\ \lambda & 1-\lambda^2
\end{pmatrix} \begin{pmatrix} a_n \\ b_n
\end{pmatrix}.
$$
Define the $2\times 2$ matrix $$M_{\lambda} = \begin{pmatrix} 1 & -\lambda \\ \lambda & 1-\lambda^2 \end{pmatrix}.$$ Then starting from a given point $(a_0, b_0)$ we find $$\begin{pmatrix} a_n \\ b_n
\end{pmatrix} = M_{\lambda}^n \begin{pmatrix} a_0 \\ b_0
\end{pmatrix}.$$
Consider the quadratic polynomial $P_\lambda(x, y) = x^2+y^2 - \lambda\, x y$. A straightforward computation shows that this polynomial is invariant under the substitution
$$\begin{pmatrix} x \\ y \end{pmatrix} \leftarrow M_{\lambda} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x - \lambda\,y \\ \lambda\,x+(1-\lambda^2)\,y \end{pmatrix}.
$$
This means that for all $n \in \mathbb{N}$ we have the equality $P_{\lambda}(a_n,b_n) = P_{\lambda}(a_0, b_0)$. In other words, all points $(a_n,b_n)$ lie on the same level set of $P_{\lambda}$. For $\lambda=0.069$ and starting point $(a_0, b_0) = (1, 10)$ this means that for all $n \in \mathbb{N}$ $$a_n^2+b_n^2 - 0.069 a_n b_n = 100.31.$$
In this case $P_{0.069}(x,y)=100.31$ defines an ellipse. For $\lambda=0$ a non-zero level set of $P_{\lambda}$ is a circle (though the sequence is stationary in this case so not very exciting). For $|\lambda|<2$ a level set is an ellipse, for $\lambda = \pm 2$ it is a pair of parallel lines and for $|\lambda|>2$ it is a hyperbola.
The recurrence relation has a nice solution in closed form. Take $\lambda \in (-2,2)$ so that the points follow an elliptical orbit. Let the parameter $s$ be such that $\sin(s) = \lambda/2$. Define
$$
\begin{eqnarray}
a_n &=& \cos(2 s \,n) \\
b_n &=& \sin(2 s \,n + s).
\end{eqnarray}
$$
One can verify that $a_n, b_n$ satisfy the recurrence relation for the parameter $\lambda$. Any solution of the recurrence with any starting point can be obtained from this fundamental solution by a change of phase and amplitude. See here for an approximation for $\lambda=0.069$ and $(a_0,b_0) = (1,10)$.
Recursions of degree $3$ are necessary (and probably sufficient), for example, if $$a_0=b_0=1,\qquad a_{n+1}=ua_n+1,\qquad b_{n+1}=vb_n+1,$$ then (I believe that) the sequence $c_n=a_nb_n$ solves the recursion
$$
c_{n+3}=(uv+u+v)c_{n+2}-uv(u+v+1)c_{n+1}+u^2v^2c_n+1-uv.$$
Best Answer
Although it is possible to solve selected non-linear recurrence relations if you happen to be lucky, in general all sorts of peculiar and difficult-to-characterize things can happen.
One example is found in chaotic systems. These are hypersensitive to initial conditions, meaning that the behavior after many iterations is extremely sensitive to tiny variations in the initial conditions, and thus any formula expressing the relationship will grow impossibly large. These recurrence equations can be amazingly simple, with xn+1 = 4xn(1-xn) with x0 between 0 and 1 as one of the classic simple examples (i.e. merely quadratic; this is the logistic map).
User @Did has already given the Mandelbrot set example--similarly simple to express, and similarly difficult to characterize analytically (e.g. by giving a finite closed-form solution).
Finally, note that to solve every non-linear recurrence relation would imply that one could solve the Halting problem, since one could encode a program as initial states and the workings of the Turing machine as the recurrence relations. So it is certainly hopeless in the most general case. (Which highly restricted cases admit solutions is still an interesting question.)