$x=f(t): af(t)+f(f(t))=(a+1)f(t)+t=f(f(f(t))-f(t))=f(t) \to f(t)=-t/a$ $0=f(f(x))-f(x)-x=x/a^2+x/a-x=x/a^2(1+a-a^2) \to a^2-a-1=0 \to a= \frac{1\pm \sqrt{5}}{2}$ $0=f(f(x)-x)-f(x)-ax=x/a^2+x/a+x/a-ax=x/a^2(1+2a-a^3) \to a^3-2a-1=0 \to a= \frac{1\pm \sqrt{5}}{2},-1$ Answer: $ a= \frac{1\pm \sqrt{5}}{2} $

Put to the second equation $x:=f(x)$. Then we get by the first equation $f(f(x))+af(x)=f(f(f(x))-f(x))=f(x)=f(f(x))-x$. So $f(x)=\frac{-x}{a}$. Putting this into the first equation, after obvious calculations we have $\frac{x}{{a}^{2}}=f(f(x))=f(x)+x=\frac{-x}{a}+x$, which gives $ a= \frac{1\pm \sqrt{5}}{2} $ when $x\neq0$. After verification of such $a$ in second equation we get that indeed $ a= \frac{1\pm \sqrt{5}}{2} $.

Interesting problem. Define $P(x) \equiv f(f(x)) = f(x) + x\ \forall x\in\mathbb{R}$. Define $Q(x) \equiv f(f(x) - x) = f(x) + ax\ \forall x\in\mathbb{R}$. Assuming $a \neq 0$. \begin{align*} Q(f(x)) &\implies f(f(f(x)) - f(x)) = f(f(x)) + af(x)\\ &\implies af(x) + x = 0 \implies f(x) = -\frac{x}{a} \end{align*}Assuming $a = 0$, $Q(f(x)) \implies x = 0\ \forall x\in\mathbb{R}$ which is absurd \[P(1) \implies \frac{1}{a^2} = - \frac{1}{a} + 1\implies a = \frac{1\pm\sqrt5}{2}\]It is simple to check this.

substitute $f(x)$ in place of $x$ in the last equation.Then,$f(f(f(x))-f(x))=f(f(x))+af(x) \implies f(x)=f(x)+x+af(x) \implies f(x)=\frac{-x}{a}$.Now,substitute it in the first equation to get, $a=\frac{1 \pm \sqrt(5)}{2}$

Let $x = f(x)$ in (ii) we have $f(x) = f(x) + x + af(x)$ or $f(x) = \frac{-x}{a}$ Now from (i) we have $\frac{x}{a^2} = \frac{-x}{a} + x$ or $a^2 - a - 1 = 0$ so $a = \frac{1\pm \sqrt{5}}{2}$.