EDIT:
The simple argument contains an error which is revealed by studying the initial value problem
\begin{align}
x(0) &= 1 \\ x'(t) &= x(t)
\end{align}
In this case, all quantities can be computed explicitly and compared. The exact solution is $$x(t) = e^t.$$ Euler's method takes the form
\begin{align} y_0 &= 1 \\ y_{j+1} &= y_j + y_j h. \end{align}
It follows that $$y_n = (1+h)^n.$$
The truncation error at $t_j = jh$ is
$$ \epsilon_j = x(t_{j+1}) - \left( x(t_j) + x(t_j)h\right) = e^{(j+1)h} - e^{jh}(1+h).$$
It follows that the sum of the truncation errors are
$$ T_n = \sum_{j=0}^{n-1} \epsilon_j = e^h \sum_{j=0}^{n-1} (e^h)^j - (1+h) \sum_{j=0}^{n-1} (e^h)^j = (e^h - 1 - h) \frac{e^{nh}-1}{e^h-1}.$$
In contrast, the global error at $t=nh$ is
$$ E_n = e^{nh} - (1+h)^n.$$
We have $T_n = E_n$ if and only if
$$ (e^{nh} - (1+h)^n)(e^h - 1) = (e^h - 1 - h)(e^{nh}-1)$$
or equivalently
$$ e^{(n+1)h} - e^{nh} - (1+h)^n e^h + (1+h)^n = e^{(n+1)h} - e^h - e^{nh} + 1-he^{nh} + h$$
or equivalently
$$(1-e^h)(1+h)^n = 1-e^{h} + h(1-e^{nh})$$
or equivalently
$$(1+h)^n = 1 + h \frac{e^{nh} - 1}{e^h - 1}$$
or equivalently
$$\frac{(1+h)^n - 1}{h} = \frac{e^{nh} - 1}{e^h - 1}$$
Now for fixed $n$ the left hand side is a polynomial in $h>0$ of degree $n-1$. The right hand side is not a polynomial of $h>0$ of degree $n-1$ unless $n=1$.
We conclude that $T_n = E_n$ if and only if $n=1$.
The simple argument contains an error which hinges on the difference between the global error and the local truncation error.
In order to eliminate any doubt, I will give all relevant definitions Consider the initial value problem
\begin{align}
x'(t) &= f(t,x(t)) \\
x(t_0) &= x_0
\end{align}
where $f$ is such that there exists a unique solution $x$ for any initial value $(t_0,x_0)$. Let $h > 0$ be given and set $t_j = t_0 + jh$. Euler's explicit method is given by
\begin{align}
y_{j+1} &= y_j + h f(t_j,y_j), \quad j = 0,1,2,\dots \\
y_0 &= x_0
\end{align}
and we use $y_j$ as an approximation of $x(t_j)$. The global error $e_j$ is given by $$e_j = x(t_j) - y_j, \quad j = 0,1,2\dotsc $$
In order to bound the global error we need an error formula. We have an iteration for $y_j$, and so it is natural to seek an iteration for $x_j:=x(t_j)$. By Taylor's formula we have at least one $\xi_j \in (t_j,t_{j+1})$, such that
$$x_{j+1} = x(t_{j+1}) = x(t_j + h) = x(t_j) + x'(t_j)h + \frac{1}{2}x''(\xi) h^2 \\= x_j + f(t_j,x_j)h + \frac{1}{2}x''(\xi_j) h^2.$$
We conclude that the $x_j$ satisfy an iteration which is quite similar to Euler's method. The local truncation error $\epsilon_j$ at time $t_{j}$ is the error term given by $$\epsilon_j = \frac{1}{2}x''(\xi_j) h^2.$$
We see that the local truncation error is the difference between $x(t_{j+1})$ and a single step of Euler's method starting from the initial point $(t_{j},x(t_j))$, i.e.,
$$ x_{j+1} - ( x_j + f(t_j,x_j)h) = \frac{1}{2} x''(\xi_j) h^2.$$
It is this relationship which allows us to derive an inequality of the type
$$ |e_{j+1}| \leq (1 + Lh) |e_j| + \frac{1}{2} M h^2,$$
which leads to the bound
$$ |e_n| \leq \frac{1}{2} Mh \frac{e^{LT} - 1}{L}, \quad 0 \leq nh \leq T$$
However, in general it is not true that the global error satisfies
$$e_n = \epsilon_0 + \epsilon_1 + \epsilon_2 + \dotsc + \epsilon_{n-1}$$
The right hand side incorporates information about single Euler steps from $n$ different points on the exact solution curve. In contrast to the left hand side, it does not have information about the repeated application of Euler's method starting from a single point.
I would write the assumptions about the error a little more explicitly, then you have less problems explaining what you do. That the error order is $a$ can be written explicitly as
$$
y_h(t)=y_E(t)+C(t)h^a+\text{higher degree terms}
$$
In the order estimate at $t=1$ the higher degree terms are disregarded, and 2 values of $h$ are used to eliminate $C(1)$ and compute $a$, 2 equations for 2 unknowns. Then you get exactly the computation you have done,
$$
\frac{y_h(1)-y_E(1)}{y_{h/2}(1)-y_E(1)}\approx 2^a
$$
from where you get the estimated value of the order $a$.
I get for a larger collection of step sizes the values
N= 1, h=1.0000, x=1.00, y= 10.54276846159383, a= 1.91118
N= 2, h=0.5000, x=1.00, y= 5.83159536458917, a= 2.10154
N= 4, h=0.2500, x=1.00, y= 4.52294038463724, a= 2.08881
N= 5, h=0.2000, x=1.00, y= 4.37414754577927, a= 2.07703
N=10, h=0.1000, x=1.00, y= 4.18433504965781, a= 2.04429
If you do the same for the explicit midpoint method (=improved Euler), then conversely the numerical values are better for the larger step sizes, but the order estimate converges slower towards $2$.
N= 1, h=1.0000, x=1.00, y= 4.48168907033806, a= 1.22604
N= 2, h=0.5000, x=1.00, y= 4.27769565474792, a= 1.55472
N= 4, h=0.2500, x=1.00, y= 4.17722465081606, a= 1.81270
N= 5, h=0.2000, x=1.00, y= 4.16044848001507, a= 1.86043
N=10, h=0.1000, x=1.00, y= 4.13503447502597, a= 1.94137
Best Answer
Example
Suppose the equation $y' = -\frac12 y$ with initial condition $y(0) = 1$ with the Euler method on the interval $x \in [0,1]$. We actually know the exact solution in this case, $y(x) = \exp(\frac12x)$.
Finding L
We note that $f(x,y) = \frac12 y$ in the example, so the equation
$$|f(x,y_1) - f(x,y_2)| \le L|y_1 - y_2|$$
becomes $$|\frac12 y_1 - \frac12 y_2| \le L |y_1 - y_2|.$$
This equation is satisfied for $L = \frac12$, so this is the value of $L$ we will take (we can also have taken a bigger value for $L$, like $L=1$, in which case we'll get a result which is also valid but not as sharp).
Finding M
We need it to satisfy $|y''(x)| \le M$. Since we know the exact solution, $y(x) = \exp(\frac12x)$, we know that $y''(x) = \frac14\exp(\frac12x)$ and so $|y''(x)| \le \frac14$ on the interval $x \in [0,1]$. However, for most equations we do not know the exact solution, so let us pretend we do not. Then what we can do is the following: differentiate the differential equation $y'(x) = \frac12 y$ to get $y''(x) = \frac12 y'$, and then substitute the differential equation in it to get
$$y''(x) = \frac12 \cdot \frac12 y = \frac14y.$$
We do know that the numerical solution given by the Euler method is a decreasing sequence for this example, so $y\le1$ and thus $|y''(x)| \le \frac14$ (as we found before).
Conclusion
So, we can take $L = \frac12$ and $M = \frac14$ in the error bound.