if $f(x)-a\int_x^{x+1}f(t)~dt$ is constant, then $f(x)$ is constant or $f(x)=Ae^{bx}+B$

A very pretty problem! Under the hypotheses stated, we will conclude that $f(x) = A e^{bx}+B$ with $A, B\ge0$. The amplitude of $B$ cannot be restricted.

Since $f$ is continuous, the integral is differentiable and by the OP's calculation, $$ \frac{d}{dx}(e^{ax} f(x)) = e^{ax}(f'(x)+a f(x)) = e^{ax} a f(x+1)\ . $$ It follows $u(x)=e^{ax}f(x)$ is nonnegative and satisfies $$ u'(x) = c u(x+1) , \quad c=a e^{-a}. $$ By a simple induction argument, $u$ is infinitely differentiable with all derivatives nonnegative on $\mathbb R$. Thus its reflection $v(x)=u(-x)$ is completely monotone, meaning all its even derivatives are nonnegative and all its odd derivatives are nonpositive.

By Bernstein's theorem on completely monotone functions, there is a unique Borel measure $d\mu$ on $[0,\infty)$ such that $$ v(x) = u(-x) = \int_0^\infty e^{-tx}\,d\mu(t) $$ for all $x>0$. But then, since $v'(x)=-u'(-x)=-c v(x-1)$, we find that for $x>1$, $$ c v(x) = -v'(x+1) = \int_0^\infty e^{-tx} e^{-t}t\,d\mu(t). $$ By uniqueness of the representing measure, $c\,d\mu(t)=te^{-t} d\mu(t)$ as measures on $[0,\infty)$. Hence $(c-te^{-t})d\mu(t)=0$, and the support of $d\mu$ must lie in the set of $t$ such that $c=te^{-t}$.

Since $t\mapsto te^{-t}$ increases on $(0,1)$ and decreases on $(1,\infty)$, this set consists of two points: the number $a\in(0,1)$ and the unique $\hat a>1$ such that $ae^{-a}=\hat a e^{-\hat a}$. Therefore $d\mu$ is a nonnegative combination of delta masses at $a$ and $\hat a$: $$ d\mu(t) = B\,\delta(t-a) + A\, \delta(t-\hat a) $$ where $B, A\ge0$. Consequently, for all $x>1$, $$ v(x) = B e^{-ax} + A e^{-\hat a x} = e^{-ax}f(-x), $$ hence $f(x) = B + A e^{bx}$ where $b=\hat a-a>0$ satisfies $a e^b = \hat a = b+a$ as desired.

The argument above actually applies for any translate $v(x-k)$. Consequently the desired representation of $f$ holds for all $x$.