No, this is not true. There is no backward uniqueness in general.
What you need is the theory of operator semigroups, and here is a simple example.
Consider the operator $Af=f'$ in the space $X=L^2[0,1]$ with domain $$D(A)=\{f\in H^1[0,1]\,,\, f(1)=0\}.$$
Then teh solution to the initial value problem is given by $x(t)=T(t)x_0$ with $$T(t)f(s):=\begin{cases} f(t+s), t+s\leq 1,\\ 0, \quad t+s>1.\end{cases}$$
Clearly, $T(t)f=0$ for all $t>1$, but the initial value problem is well-posed in the classical sense ($A$ closed, classical solutions from a dense set of initial values, continuos dependence on initial values).