Welcome!
First of all, I suppose the proposed PDE should be
$$
F_t=axF_x+\frac{1}{2}F_{xx}.
$$
Perhaps you have missed an $x$ in front of the $F_x$ term. But please do let me know if this is not the case.
In view of $F(t-s,X_s)$, $s$ is the variable to which ${\rm d}$ is with respect, while $t$ is a parameter. Hence to be less ambiguous, let us use $F(T-t,X_t)$ instead, where $t\in\left[0,T\right]$.
Note that $F(T-t,x)$ is a function of $t$ and $x$. To make it clear, let us define $G(t,x)=F(T-t,x)$. By Ito's formula,
\begin{align}
&{\rm d}F(T-t,X_t)\\
&={\rm d}G(t,X_t)\\
&=\frac{\partial G}{\partial t}(t,X_t)\,{\rm d}t+\frac{\partial G}{\partial x}(t,X_t)\,{\rm d}X_t+\frac{1}{2}\frac{\partial^2G}{\partial x^2}(t,X_t)\,{\rm d}\left<X\right>_t.
\end{align}
Provided that
$$
{\rm d}X_t=aX_t\,{\rm d}t+{\rm d}W_t,
$$
we have ${\rm d}\left<X\right>_t={\rm d}t$. Substitute these two results into the above equation, and we obtain
\begin{align}
&{\rm d}F(T-t,X_t)\\
&=\left(\frac{\partial G}{\partial t}(t,X_t)+aX_t\frac{\partial G}{\partial x}(t,X_t)+\frac{1}{2}\frac{\partial^2G}{\partial x^2}(t,X_t)\right){\rm d}t+\frac{\partial G}{\partial x}(t,X_t)\,{\rm d}W_t\\
&=\left(\frac{\partial G}{\partial t}+ax\frac{\partial G}{\partial x}+\frac{1}{2}\frac{\partial^2G}{\partial x^2}\right)(t,X_t)\,{\rm d}t+\frac{\partial G}{\partial x}(t,X_t)\,{\rm d}W_t.
\end{align}
Recall that $G(t,x)=F(T-t,x)$, and it is obvious that
\begin{align}
\frac{\partial G}{\partial t}(t,x)&=-\frac{\partial F}{\partial t}(T-t,x),\\
\frac{\partial G}{\partial x}(t,x)&=\frac{\partial F}{\partial x}(T-t,x),\\
\frac{\partial^2G}{\partial x^2}(t,x)&=\frac{\partial^2F}{\partial x^2}(T-t,x).
\end{align}
Consequently, we obtain
\begin{align}
&{\rm d}F(T-t,X_t)\\
&=\left(-\frac{\partial F}{\partial t}+ax\frac{\partial F}{\partial x}+\frac{1}{2}\frac{\partial^2F}{\partial x^2}\right)(T-t,X_t)\,{\rm d}t+\frac{\partial F}{\partial x}(T-t,X_t)\,{\rm d}W_t.
\end{align}
Finally, note that the PDE gives
$$
-\frac{\partial F}{\partial t}+ax\frac{\partial F}{\partial x}+\frac{1}{2}\frac{\partial^2F}{\partial x^2}=0,
$$
and we eventually obtain
$$
{\rm d}F(T-t,X_t)=\frac{\partial F}{\partial x}(T-t,X_t)\,{\rm d}W_t.
$$
That's it! Hope this could be somewhat helpful for you.