Your claim holds true if and only if $-1$ is not an eigenvalue of $\Delta$.

If there is a nontrivial function $v:M\to\mathbb R$ with $\Delta v=-v$ (then necessarily $\int_Mv=0$), then $u(x,t)=v(x)$ solves your PDE with all conditions but does not vanish identically.

Suppose then that $-1$ is not an eigenvalue.
Then it follows that the line $-1+i\mathbb R\subset\mathbb C$ is in the resolvent set of $\Delta$.
Because of the boundary condition the solution $u$ can be thought of as a periodic solution (w.r.t. time).
Then it is very convenient to take the Fourier series of $u$ in $t$:
$$
\hat u(x,s)
=
\int_0^Tu(x,t)e^{2\pi its/T}dt
$$
for $s\in\mathbb Z$.
Since $\widehat{u_t}=-\frac{2\pi is}{T}\hat u$, the PDE becomes
$$
\left(\Delta+(1+\frac{2\pi is}{T})\right)\hat u
=
0.
$$
Notice that the equations for different Fourier components decouple.
Since $-(1+\frac{2\pi is}{T})$ is in the resolvent for all $s$, we have $\hat u=0$ and consequently $u=0$.

Even if we do not assume that $-1$ is not an eigenvalue, we know that the spectrum is real.
Therefore the above argument shows that any solutions $u$ has to be constant in time ($\hat u(x,s)=0$ when $s\neq0$) and thus an eigenfunction of the Laplacian — and the eigenvalue should come as no surprise.