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
Let $k=\lfloor x\rfloor$
$$\prod\limits^{\lfloor x\rfloor}_{n=1}\frac{\lceil x\rceil}{\lfloor x\rfloor}=\prod\limits^{k}_{n=1}\frac{k+1}{k}=\left( 1+\frac1k\right)^k$$
therefore,
$$\lim_{x\to\infty\land x\notin\mathbb N}\prod\limits^{\lfloor x\rfloor}_{n=1}\frac{\lceil x\rceil}{\lfloor x\rfloor}=\lim_{k\to\infty}\left( 1+\frac1k\right)^k=e$$