Your computation of eigenvalues is wrong. Of the system matrix in $y'=Ay$ there is one eigenvalue $-200$ with eigenvector $\pmatrix{1\\1}$ and one eigenvalue $-1000$ with eigenvector $\pmatrix{1\\-1}$. These translate into eigenvalues $1-200h$ and $1-1000h$ of the "driving matrix" for the Euler method, requiring $h<0.002$ for stability.
For the implicit method there are no step size restrictions to get stability in the method, to get into the range where the error behaves like order 1 one still will need $500h\ll 1$.
In this specific case you can find the explicit formula for $T$: $u(t)=e^{At}u_{0}$. Here $e^{At}$ is defined in terms of usual Taylor expansion of the exponential function, evaluated at the matrix $At$. By Picard-Lindelof theorem the solution $u(t)$ must be uniquely given, and the particular choice $u(t)=e^{At}u_{0}$ satisfies the equation so it must be the unique solution. (Or, you can directly show it must be the solution by verifying that $e^{-At}u(t)$ must have zero derivative, so must be a constant.)
Therefore, in your equation
$$
\left\langle T^{*}v,u_{0} \right\rangle_{2}
= \int_{0}^{1}v(t)\cdot u(t)\,dt,
$$
the right-hand side is equal to
$$
\int_{0}^{1} v(t)\cdot e^{At}u_{0}\,dt = \left(\int_{0}^{1}e^{A^{*}t}v(t)\,dt\right)\cdot u_{0}
$$
(here we used the identity $(e^{At})^{*}=e^{A^{*}t}$), thus you conclude
$$
T^{*}v = \int_{0}^{1}e^{A^{*}t}v(t)\,dt.
$$
Probably it is also possible to obtain a similar formula when the ODE is of more general form. Not so sure what precisely such a statement would be.
Edit
Regarding your comment. Consider the ODE
\begin{align*}
w'(t) &= -A^{*}w(t) - v(t), \\
w(1) &= 0.
\end{align*}
Then with some similar computations, you can see that the solution is given as
$$
w(t) = e^{-A^{*}t}\int_{t}^{1}e^{A^{*}\tau}v(\tau)\,d\tau.
$$
Then $T^{*}v$ is precisely $w(0)$. In other words, you are considering the linear ODE with the coefficient matrix $A^{*}$ instead of $A$, while the input $v$ acts as the "forcing term", and you are solving the ODE "backward", so you are fixing the "final value" instead of the "initial value". The minus signs in the ODE also accounts for the direction of time being backward. Then the output of $T^{*}$ is the initial value of the ODE.
(The "general case" I mentioned in the end of the answer before the edit probably would be something similar.)
Now $T^{*}$ being given in terms of the solution $w$ of the ODE above can be derived directly as follows, which is probably what you want. Since
$$
v(t) = -A^{*}w(t) - w'(t),
$$
we can write
\begin{align*}
\left\langle T^{*}v,u_{0} \right\rangle_{2}
&= -\int_{0}^{1}A^{*}w(t)\cdot u(t) + w'(t)\cdot u(t)\,dt \\
&= -\int_{0}^{1}w(t)\cdot Au(t) + w'(t)\cdot u(t)\,dt \\
&= -\int_{0}^{1}(w(t)\cdot u(t))'dt \\
&= -(w(1)\cdot u(1) - w(0)\cdot u(0)) \\
&= w(0)\cdot u_{0},
\end{align*}
thus we get $T^{*}v = w(0)$.
Best Answer
Note that in general to get to barely useful results with the Euler methods, one should observe the guideline that $Lh<1.5$. Here with $L\approx 2$ and $h=0.5$ that is satisfied.
The explicit step should have $u_0(2.5)=0.5$, $u_1(0.5)=1+0.5⋅2=2$ and $u_2(2.5)=2+0.5⋅(2⋅0-1)=1.5$, somehow in the omitted part there are some terms missing.
For the implicit step, you get a non-trivial system $u(t)=(I-hA)u(t+h)$ that you have to solve via Gauß elimination or similar. $$ \pmatrix{0\\1\\2} = \pmatrix{1&-0.5&0\\0&1&-0.5\\-0.5\cdot 2.5&0.5&1} \pmatrix{u_0(2.5)\\u_1(2.5)\\u_2(2.5)} $$ which solves as $u_0(2.5)=1.06666667$, $u_1(2.5)=2.13333333$, $u_2(2.5)=2.26666667$.
A more exact integration with a higher order method gives the value table \begin{array}{c|ccc} t&u_0(t)&u_1(t)&u_2(t)\\\hline 2.000 & 0.00000000 & 1.00000000 & 2.00000000 \\ 2.100 & 0.10983395 & 1.19502947 & 1.90118241 \\ 2.200 & 0.23868570 & 1.38047638 & 1.80957168 \\ 2.300 & 0.38564537 & 1.55743288 & 1.73264001 \\ 2.400 & 0.54995008 & 1.72774964 & 1.67807126 \\ 2.500 & 0.73106066 & 1.89405346 & 1.65369605 \\ \end{array} which is not exactly close in the last row.