Suppose
$y(0) = y_0; \tag{1}$
then the unique solution to the equation
$y' = a(x)y + b(x) \tag{2}$
is
$y(x) = \exp \left ( \displaystyle \int_0^x a(s)ds \right ) \left (y_0 + \displaystyle \int_0^x \exp \left (-\displaystyle \int_0^s a(u) du \right )b(s) ds \right ); \tag{3}$
formula (3) is very well known; a derivation may be found here; for $x = T$ we thus have
$y(T) = \exp \left ( \displaystyle \int_0^T a(s)ds \right ) \left (y_0 + \displaystyle \int_0^T \exp \left (-\displaystyle \int_0^s a(u) du \right )b(s) ds \right ). \tag{4}$
A $T$-periodic solution to (2) satisfies $y(T) = y(0) = y_0$; in this case (4) yields
$y_0 = \exp \left ( \displaystyle \int_0^T a(s)ds \right ) \left (y_0 + \displaystyle \int_0^T \exp \left (-\displaystyle \int_0^s a(u) du \right )b(s) ds \right ), \tag{5}$
which we re-write as
$\left ( 1 - \exp \left ( \displaystyle \int_0^T a(s) ds\right)\right)y_0 = \exp \left ( \displaystyle \int_0^T a(s) ds \right )\displaystyle \int_0^T \exp \left (-\displaystyle \int_0^s a(u) du\right) b(s) ds; \tag{6}$
the hypothesis
$\displaystyle \int_0^T a(s) ds \ne 0 \tag{7}$
guarantees that
$\exp \left ( \displaystyle \int_0^T a(s) ds \right ) \ne 1; \tag{8}$
in this case we may solve (6) for $y_0$:
$y_0 = \dfrac{\displaystyle \int_0^T a(s) ds}{1 - \exp \left (\displaystyle \int_0^T a(s) ds \right )}\displaystyle \int_0^T \exp \left ( \displaystyle \int_0^s a(u) du \right ) b(s) ds. \tag{9}$
(9) indicates that there is at most one initial condition $y_0 = y(0)$ for which the solution $y(t)$ of (2) is periodic. Thus any periodic $y(t)$ satisfying (2) under the condition (7) must be unique; it remains to establish the existence of such a $y(t)$.
To establish the existence of a periodic solution, note that we may translate any solution to (2) forward in $x$ by $T$, obtaining a function $y(x + T)$; we have
$y'(x +T) = a(x + T)y(x + T) + b(x + T) = a(x)y(x + T) + b(x), \tag{10}$
by the $T$-periodicity of $a(x)$ and $b(x)$. Furthermore, at $x = 0$ the function $y(x + T)$ takes the value
$y(0 + T) = y(T). \tag{11}$
If follows from (11) that if we can find a solution $y(x)$ such that
$y(T) = y(0) = y_0, \tag{12}$
then both $y(x)$ and $y(x + T)$ will satisfy (2) with the same initial condition $y_0$; uniqueness of solutions then allows us to conclude that
$y(x + T) = y(x) \tag{13}$
for all $x$; i.e., the solution $y(x)$ is $T$-periodic. But if we choose $y_0$ as in (9), then (6) and hence (5) evidently bind, so that (4) yields (12), and hence (13); we have thus demonstrated the existence of a periodic solution to (2), which must then be unique by the arguments given above. QED.
Best Answer
This is wrong. The counterexample is $$ A=\left(\begin{array}{cc}0&2\pi\\-2\pi &0\end{array}\right),\qquad e^A=I. $$ Here $A$ has the eigenvalues $\pm2\pi i$.
Let $u(t)$ be a nontrivial periodic solution with period $p$ and prove that there exists some $s$, $s\in (0,p)$, such that the vectors $u(0)$ and $u(s)$ are linearly independent. Suppose the opposite is true, i.e. $$ \forall s\in [0,p]\; \exists c(s)\in\mathbb R\, :\; u(s)= c(s)u(0). $$ It means that the periodic solution lays on the straight line determined by the points $0$ and $u(0)$, which is impossible due to the existence and uniqueness theorem.
Since $u(t)$ is a $p$-periodic solution, $e^{Ap}u(0)=u(0)$ and $e^{Ap}u(s)=u(s)$, hence $e^{Ap}$ has an eigenbasis $(u(0),u(s))$ and the eigenvalues are $1,1$. It implies $e^{Ap}=I$.
Indeed, it means that $e^{Ap}$ is diagonalizable: $$ U^{-1}e^{Ap} U= I\;\Rightarrow\; e^{Ap} U=U I\;\Rightarrow\; e^{Ap}=U U^{-1}=I, $$ where $U$ is a block matrix of the column vectors $(u(0),u(s))$.