When $X,Y$ are iid, the distribution of $Z=X+Y$ can be one of discrete or continuous uniform distributions, studied in the following.
For the continuous case, two cases are considered: uniformity on a single interval and on a number of disjoint intervals.
Discrete Uniform Distribution
Let the support $S_X$ of $X$ be a finite set including $a_1<\cdots<a_n$ with corresponding probabilities $p_1,\cdots,p_n>0$. For $Z$ to have a discrete uniform distribution, the following system:
$$\mathbb P(Z=2a_1)=p^2_1=\mathbb P(Z=2a_n)=p^2_n=\mathbb P(Z=a_1+a_n)=2p_1p_n+\beta, \, \beta\ge 0, $$
needs to have a non-zero solution $p_1$ and $p_n$, which is impossible (as $x^2=2x^2+\beta$ cannot have such a solution) unless $n=1$, i.e. $X$ is a constant.
Continuous Uniform Distribution on a Single Interval
Here, I provide an elementary proof (not using complex analysis) based on the inequalities given in the OP for iid $X,Y \in [0,\frac{1}{2}] $ and $\epsilon \in \left (0,\frac{1}{4} \right)$:
$$\color{blue}{P(Z\leq\epsilon)\leq F(\epsilon)^2} \tag{1}$$
$$\color{blue}{\mathbb P \left (\frac12-\epsilon\leq Z\leq \frac12+\epsilon\right )\geq2F(\epsilon)\left (1-F\left (\left [\frac12-\epsilon \right ]^- \right) \right)} \, \tag{2}.$$
To indirectly prove the statement, let us assume $Z\sim \mathcal U(0,1)$. Then, it can be shown that the mfgs of both $\frac12-X$ and $X$ are the same, so we have
$$\frac12-X\sim X,\frac12-Y\sim Y \tag{3},$$ (an alternative way to prove (3) is to show that $X-\frac{1}{4},Y-\frac{1}{4} $ have symmetric distributions by the fact that the characteristic function of any symmetric distribution is real-valued.)
Considering (3), we have
$$F(\epsilon)=\mathbb P(X \le \epsilon)=\mathbb P \left (\frac12-X \ge \frac12 -\epsilon \right )=\mathbb P \left (X \ge \frac12 -\epsilon \right)=1-F\left (\left [\frac12-\epsilon \right ]^- \right).$$
Hence, by (3) from (1) and (2) for $Z\sim \mathcal U(0,1)$, for $\epsilon \in \left (0,\frac{1}{4} \right)$ we get $\epsilon\leq F(\epsilon)^2$ and $2\epsilon\geq 2 F(\epsilon)^2$, so
$$F(\epsilon)=\begin{cases}
\sqrt{\epsilon}, & 0\le \epsilon \le \frac14\\
1-\sqrt{\frac12 - \epsilon} & \frac14 <\epsilon \le \frac12
\end{cases},$$
for which $Z \nsim \mathcal U(0,1)$, a contradiction that proves the statement.
Proofs for (1) and (2)
As $X,Y\ge 0$, we have the following implication for $\epsilon \in \left (0,\frac{1}{2} \right)$:
$$Z\leq\epsilon \Rightarrow X\leq\epsilon, Y\leq\epsilon,$$
which shows the left event is a subset of the right event, so we have (1).
Moreover, considering $0\le X,Y \le \frac{1}{2}$, we have the following implication for $\epsilon \in \left (0,\frac{1}{4} \right)$:
$$X\leq\epsilon, Y\ge \frac12-\epsilon \, \text{or}\, Y\leq\epsilon, X\ge \frac12-\epsilon\Rightarrow \frac12-\epsilon\leq Z\leq \frac12+\epsilon,$$
by noting that $Z=Y+X\leq\epsilon+Y\le \frac12+\epsilon, Z=X+Y\ge \frac12-\epsilon +Y\ge \frac12-\epsilon$. Since $\epsilon\in(0,\frac14)$, the two events on the LHS of the above implication are disjoint, and we obtain (2).
Continuous Uniform Distribution on Disjoint Intervals
By contradiction, let $Z$ have a uniform distribution over the union of two disjoint intervals $[0,a] \cup [1-b,1]$ with $0<a<1-b<1$. Then, the cdf $F$ satisfies the following equation for $0<z<a$:
$$\frac{z}{a} \frac{a}{a+b} = \int_{0}^x F(z-t)\text{d}F(t).$$
Defining $\bar{F}=\sqrt{\frac{a+b}{a}}F$ for $0<z<a$, this can be written as
$$\frac{z}{a}= \int_{0}^x \bar{F}(z-t)\text{d}\bar{F}(t),$$
which means that the sum of iid $X,Y$ with the cdf $\bar{F}$ has a uniform distribution on $(0,a)$, already shown not possible in the previous section, and we get a contradiction.
Unsolved Special Case: An Infinite Sequence of Disjoint Intervals with Unbounded Below Union
The only case that I could not yet find an answer for it is the following case (hope readers can help to solve this case too). Suppose that the support of $Z$ is not bounded from below and includes disjoint intervals $(l_i,l_i+\delta_i), i\in \mathbb Z$ with $\Delta=\sum_{i\in \mathbb Z}\delta_i<\infty $. Note that if $\inf_{i\in \mathbb Z} \{l_i \}>-\infty$, we can handle it using the method used in the previous section.
