Update for clarification:
Assume $\dot{x_1}=f(x_1 , x_2)+ au$
where $x_1$ is asymptotically stable for all bounded values of $x_2$. If $x_2$ is kept bounded, will input-output stability theorem apply for a system with $y=x_1$.
Khalil's theorem #5.1 and #5.2 consider systems that need to have asymptotically stable unforced version ($u=0$).
Are there any theorems for a system that its unforced version is neutrally stable? For example:
\begin{cases} \dot{x_1}=f(x_1 , x_2)+ au & \text{}\\ \dot{x_2}=bu & \text{} \end{cases}
where the unforced system is only neutrally stable.
Can we show small-gain stability and bounds for the output $ y=x_1$ assuming $u$ is $L_2$ bounded, for example?
I don't think it is possible to conclude anything about stability of such a system without further assumptions because you can find counterexamples even if you consider linear systems only. Take for example
$$ \begin{align} \dot{x}_1&=x_2\\ \dot{x}_2&=-x_1+u\\ y&=x_1 \end{align} $$
which is neutrally stable since the eigenvalues of its $A$ matrix are $\pm j$. However take $$ u=\frac{\sin(t)}{t + 1} $$ and zero initial conditions. Note that $u$ is bounded and smooth on $t\in[0,\infty)$. It is also in $L_2$, as
$$ \begin{align} \Vert u\Vert_{L_2}^2 &= \int_0^{\infty}u(\sigma)^Tu(\sigma)\text{d}\sigma \\ &= \text{Ci}(2)\sin(2) - \text{Si}(2)\cos(2) + \frac{\pi}{2}\cos(2) \\ &\approx 0.399021 < \infty \end{align} $$
which can be verified with Wolfram Alpha. Here $\text{Ci}$ is the cosine integral function and $\text{Si}$ the sine integral function.
Using again Wolfram Alpha, the explicit solution is computed as $$ y(t)= y_u(t) + y_b(t) $$ with $$ \begin{align} y_u(t)&=-\frac{1}{2} \log(t + 1) \cos(t) \\ y_b(t)&= c_2 \sin(t) + c_1 \cos(t) + \frac{1}{2} \text{Ci}(2 t + 2) \cos(t + 2) + \frac{1}{2} \text{Si}(2 t + 2) \sin(t + 2) \end{align} $$ with constants $c_1,c_2$. Note that $y_b(t)$ is bounded because $\text{Ci}$ and $\text{Si}$ are bounded on $[2,\infty)$.
However, $y_u(t)$ is unbounded as the logarithm (although slow) keeps growing forever, so $y(t)$ grows (oscillating) above all bounds.