We want to prove that
the unique solution to the following difference equation is the null one:
$$
au(x)+b\mathbf{1}\_{(0,\frac{1}{2})}(x)u(x+\frac{1}{2})+c\mathbf{1}\_{(\frac{1%
}{2},1)}(x)u(x-\frac{1}{2})=0,\text{ }x\in (0,1).
$$
Extending $u$ by zero outside $(0,1)$ and taking the Laplace transform yields
$$
a\int\_{0}^{1}e^{-px}u(x)dx+be^{\frac{p}{2}}\int\_{\frac{1}{2}%
}^{1}e^{-px}u(x)dx+ce^{-\frac{p}{2}}\int\_{0}^{\frac{1}{2}}e^{-px}u(x)dx=0,%
\text{ }p\in 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
,
$$
that is
$$
\left( a+be^{\frac{p}{2}}\right) \int\_{\frac{1}{2}}^{1}e^{-px}u(x)dx+\left(
a+ce^{-\frac{p}{2}}\right) \int\_{0}^{\frac{1}{2}}e^{-px}u(x)dx=0,\text{ }%
p\in 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
.
$$
If we let for instance $p=\gamma +4n\pi i,n\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ with $a+ce^{-\frac{\gamma }{2}}=0$ we get
$$
\left( a+be^{\frac{\gamma }{2}}\right) e^{-\gamma }\int\_{\frac{1}{2}%
}^{1}e^{-4n\pi ix}u(x)dx=0,
$$
which yields that $u=0$ on $(\frac{1}{2},1)$ if $a+be^{\frac{\gamma }{2}%
}\neq 0$ which is equivalent to $a^2-bc \neq 0$. With the same manner, by choosing this time $p=\delta +4n\pi
i,n\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ with $a+be^{\frac{\delta }{2}}=0$ we get
$$
\left( a+ce^{-\frac{\delta }{2}}\right) e^{-\delta }\int\_{0}^{\frac{1}{2}%
}e^{-4n\pi ix}u(x)dx=0,
$$
so if $a+ce^{-\frac{\delta }{2}}\neq 0$ we get $u=0$ on $(0,\frac{1}{2})$, which is equivalent to $a^2-bc \neq 0$.


I don't know if this kind of reasoning is correct since the Laplace
transform of $u$ is zero on subintervals with different choices of $p.$



Unique zero solution to a difference equation via Laplace transform
Your condition can be rewritten as the system of three equations:  

$$au(x)+bu(x+1/2)=0\ \forall x\in(0,1/2), \tag{1}$$
$$au(x)+cu(x-1/2)=0\ \forall x\in(1/2,1), \tag{2} $$
$$au(1/2)=0. \tag{3} $$
In turn, (2) can be rewritten as
$$cu(x)+au(x+1/2)=0\ \forall x\in(0,1/2). \tag{2a} $$
So, if the determinant $a^2-bc$ of the system (1)--(2a) of linear equations for $u(x),u(x+1/2)$ is nonzero, then $u(x)=u(x+1/2)=0$ $\forall x\in(0,1/2)$, that is, $u(x)=0$ $\forall x\in(0,1)\setminus\{1/2\}$. Together with (3), this yields $u(x)=0$ $\forall x\in(0,1)$ if $a\ne0$.


The cases when $a^2-bc=0$ or $a=0$ are considered similarly.


In particular, if $a^2-bc=0$ but $a\ne0$, then $a\ne0$ and $b\ne0$, then the system (1)--(3) reduces to the conditions
$$u(x)=-(b/a)u(x+1/2)\ \forall x\in(0,1/2)\tag{1a}$$ and $u(1/2)=0$. So, here one may assign any values to $u$ on $(1/2,1)$, and then use (1a) to determine the values of $u$ on $(1/2,1)$.




---


The OP requested in a comment that the solution be given in terms of the Laplace transform, say $L$. This can be done as follows.


Of course, to use the Laplace transform, we have to assume that $u$ is integrable on $(0,1)$. Let
$$U(x):=\begin{cases}u(x)&\text{ if }0<x<1/2,\\ 
0&\text{ if }x>1/2,\end{cases}$$
$$V(x):=\begin{cases}u(x+1/2)&\text{ if }0<x<1/2,\\ 
0&\text{ if }x>1/2.\end{cases}$$
Then (1) and (2) imply
$$aL(U)+bL(V)=0,$$
$$cL(U)+aL(V)=0.$$
So, if $a^2-bc\ne0$, then $L(U)=L(V)=0$ and hence $U=V=0$ almost everywhere (a.e.), so that $u=0$ a.e., so that $u=0$ if $u$ is continuous.


As shown above, there is no uniqueness if $a^2-bc=0$.


