Your original statement is not true (please read the whole answer). But the following statement is correct:
$$\left \lfloor {\log_2 n} \right \rfloor = \left \lfloor {\log_2 \left \lceil \frac {n-1}{2}\right \rceil} \right \rfloor + 1\tag{1}$$
Proof:
Every number $n$ can be placed between two cosecutive powers of 2. In other words, there exists $k$ such that:
$$2^k\le n\le2^{k+1}-1\tag{1}$$
Obviously:
$$k\le\log_2n<k+1$$
$$k=\lfloor\log_2n\rfloor\tag{2}$$
On the other side from (1):
$$\frac{2^k-1}2 \le \frac{n-1}{2} \le \frac{2^{k+1}-2}2$$
$$2^{k-1}-\frac12 \le \frac{n-1}{2} \le 2^k-1$$
$$\lceil 2^{k-1}-\frac12\rceil \le \lceil\frac{n-1}{2}\rceil \le 2^k-1$$
$$2^{k-1} \le \lceil\frac{n-1}{2}\rceil \le 2^k-1$$
$${k-1} \le \log_2\lceil\frac{n-1}{2}\rceil \le \log_2 (2^k-1)<k$$
$${k-1} =\lfloor \log_2\lceil\frac{n-1}{2}\rceil \rfloor$$
$$k=\lfloor \log_2\lceil\frac{n-1}{2}\rceil \rfloor+1\tag{3}$$
By comparing (2) and (3) you get:
$$\lfloor\log_2n\rfloor = \lfloor \log_2\lceil\frac{n-1}{2}\rceil \rfloor+1$$
...which completes the proof.
You can easily prove that the original statement is not true. You are basically saying that the function:
$$f(n)=\left \lfloor {\log n} \right \rfloor - \left \lfloor {\log \left \lceil \frac {n-1}{2}\right \rceil} \right \rfloor - 1$$
...is equal to zero for all values of $n$.
This is not true if "$\log$" stands for logartihm with base 10:
This is also not true if "$\log$" stans for natural logarithm "$\ln$":
If you don't trust these plots, calculate the value for $n=45$ and in both cases the result is -1, not 0.
Best Answer
As mentioned in comment, the first two conditions $T(1) = 0, T(2) = 1$ is incompatible with the last condition $$\require{cancel} T(n) = T(\lfloor\frac{n}{2}\rfloor) + T(\lceil\frac{n}{2}\rceil) = 2\quad\text{ for } \color{red}{\cancelto{\;\color{black}{n > 2}\;}{\color{grey}{n \ge 2}}} \tag{*1}$$ at $n = 2$. We will assume the condition $(*1)$ is only valid for $n > 2$ instead.
Let $\displaystyle\;f(z) = \sum\limits_{n=2}^\infty T(n) z^n\;$ be the generating function for the sequence $T(n)$. Multiply the $n^{th}$ term of $(*1)$ by $z^n$ and start summing from $n = 3$, we obtain:
$$\begin{array}{rrl} &f(z) - z^2 &= T(3) z^3 + T(4) z^4 + T(5) z^5 + \cdots\\ &&= (T(2) + 2)z^3 + (T(2) + T(2) + 2)z^4 + (T(2)+T(3)+2)z^5 + \cdots\\ &&= (1+z)^2 ( T(2)z^3 + T(3)z^5 + \cdots) + 2(z^3 + z^4 + z^5 + \cdots)\\ &&= \frac{(1+z)^2}{z}f(z^2) + \frac{2z^3}{1-z}\\ \implies & f(z) &= \frac{(1+z)^2}{z}f(z^2) + z^2\left(\frac{1+z}{1-z}\right)\\ \implies & \frac{(1-z)^2}{z} f(z) &= \frac{(1-z^2)^2}{z^2}f(z^2) + z(1-z^2) \end{array} $$ Substitute $z^{2^k}$ for $z$ in last expression and sum over $k$, we obtain
$$f(z) = \frac{z}{(1-z)^2}\sum_{k=0}^\infty \left(z^{2^k} - z^{3\cdot2^k}\right) = \left( \sum_{m=1}^\infty m z^m \right)\sum_{k=0}^\infty \left(z^{2^k} - z^{3\cdot2^k}\right)$$
With this expression, we can read off $T(n)$ as the coefficient of $z^n$ in $f(z)$ and get
$$T(n) = \sum_{k=0}^{\lfloor \log_2 n\rfloor} ( n - 2^k ) - \sum_{k=0}^{\lfloor \log_2(n/3)\rfloor} (n - 3\cdot 2^k)$$
For $n > 2$, we can simplify this as $$\bbox[4pt,border: 1px solid black;]{ T(n) = n \color{red}{\big(\lfloor \log_2 n\rfloor - \lfloor \log_2(n/3)\rfloor\big)} - \color{blue}{\big( 2^{\lfloor \log_2 n\rfloor + 1} - 1 \big)} + 3\color{blue}{\big( 2^{\lfloor \log_2(n/3)\rfloor +1} - 1\big)}}\tag{*2}$$
There are several observations we can make.
When $n = 2^k, k > 1$, we have $$T(n) = n(k - (k-2)) - (2^{k+1} - 1) + 3(2^{k-1} - 1) = \frac32 n - 2$$
When $n = 3\cdot 2^{k-1}, k > 0$, we have $$T(n) = n(k - (k-1)) - (2^{k+1} - 1) + 3(2^k - 1) = \frac53 n - 2$$
For $2^k < n < 3\cdot 2^{k-1}, k > 1$, the coefficient for $n$ in $(*2)$ (i.e the factor in red color) is $2$, while the rest (i.e those in blue color) didn't change with $k$. So $T(n)$ is linear there with slope $2$.
Combine these, we find in general
$$\frac32 n - 2 \le T(n) \le \frac53 n - 2 \quad\text{ for }\quad n > 2$$ and $T(n) = O(n)$ as expected. However, $\frac{T(n)}{n}$ doesn't converge to any number but oscillate "between" $\frac32$ and $\frac53$.
Above is a picture ilustrating the behavior of $T(n)$. The blue pluses are the value of $T(n) - (\frac32 n - 2)$ computed for various $n$. The red line is $\frac{n}{6} = (\frac53 n - 2) - (\frac32 n - 2)$. As one can see, $T(n)$ doesn't converge to any straight line. Instead, it oscillate between lines $\frac32 n - 2$ and $\frac53 n - 2$ as discussed before.