Is this Riemann zeta function product equal to the Fourier transform of the von Mangoldt function?

Your first two relations follow quickly from the asymptotic behavior of the functions $s\mapsto\zeta(s)$ and $s\mapsto d^{1-s}$ as $s\to 1$. Let me demonstrate this for the second relation for $n>1$ (note that it fails for $n=1$).

For $|s-1|<1$ we have $$\zeta(s)=1/(s-1)+O(1)\quad\text{and}\quad d^{1-s}=1-(s-1)\log d+O_d((s-1)^2),$$ hence $$ \sum_{d\mid n}\frac{\mu(d)}{d^{s-1}} =\sum_{d\mid n}\mu(d)-(s-1)\sum_{d\mid n}\mu(d)\log d+O_n((s-1)^2).$$ On the right hand side the first sum is zero, and the second sum is $-\Lambda(n)$, because the convolution of the Möbius function with $1$ and $\log$ at $n>1$ equals $0$ and $\Lambda(n)$, respectively. It follows that $$ \sum_{d\mid n}\frac{\mu(d)}{d^{s-1}} \sim (s-1)\Lambda(n)\quad\text{as}\quad s\to 1, $$ whence $$ \lim_{s\to 1}\zeta(s)\sum_{d\mid n}\frac{\mu(d)}{d^{s-1}} =\Lambda(n).$$

Now I don't understand your main question. What do you mean by the Fourier transform of $\Lambda(n)$? The Fourier transform is initially defined for sequences in $L^1(\mathbb{Z})$, and then by a limit procedure for sequences in $L^2(\mathbb{Z})$. However, $\Lambda(n)$ is not even bounded. Please define precisely the left hand side of your third display, and explain what you mean by the symbol $\sim$ there.

P.S. Similarly, I don't understand what you mean by $f(t)$. You define it by an infinite series, but I don't see that the series converges.

This is not a complete answer but I want show that there is nothing mysterious going on here.

We want to prove that:

$$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$

The Dirichlet inverse of the Euler totient function is

$$a(n)=\sum\limits_{d|n} \mu(d)d$$

Construct the matrix $$T(n,k)=a(GCD(n,k))$$

which starts:

$$\displaystyle T = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{bmatrix} $$

where GCD is the Greatest Common Divisor of row index $n$ and column index $k$.

joriki showed that the von Mangoldt function is $$\Lambda(n)=\sum\limits_{k=1}^{k=\infty} \frac{T(n,k)}{k}$$

Then add this quote by Terence Tao from here, that I don't completely understand but I do almost see why it should be true:

Quote:" The Fourier transform in this context becomes (essentially) the Mellin transform, which is particularly important in analytic number theory. (For instance, the Riemann zeta function is essentially the Mellin transform of the Dirac comb on the natural numbers"

Now let us return to the matrix $T$:

First the von Mangoldt is expanded as:

$$\displaystyle \begin{bmatrix} +1/1&+1/1&+1/1&+1/1&+1/1&+1/1&+1/1&\cdots \\ +1/2&-1/2&+1/2&-1/2&+1/2&-1/2&+1/2 \\ +1/3&+1/3&-2/3&+1/3&+1/3&-2/3&+1/3 \\ +1/4&-1/4&+1/4&-1/4&+1/4&-1/4&+1/4 \\ +1/5&+1/5&+1/5&+1/5&-4/5&+1/5&+1/5 \\ +1/6&-1/6&-2/6&-1/6&+1/6&+2/6&+1/6 \\ +1/7&+1/7&+1/7&+1/7&+1/7&+1/7&-6/7 \\ \vdots&&&&&&&\ddots \end{bmatrix}$$

Edit: 24.1.2016. From here on the variables $n$ and $k$ should be permutated but I don't know how to fix the rest of this answer right now.

Summing the columns first is equivalent to what was said earlier: $$\Lambda(n)=\sum\limits_{k=1}^{k=\infty} \frac{T(n,k)}{k}$$

where:

$$\Lambda(n) = \begin{cases} \infty & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

or as a sequence:

$$\infty ,\log (2),\log (3),\log (2),\log (5),0,\log (7),\log (2),\log (3),0,\log (11),0,...,\Lambda(\infty)$$

And now based on the quote above let us say that: $$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) = \sum\limits_{n=1}^{n=k}\frac{\Lambda(n)}{n^s}$$

Expanding this into matrix form we have the matrix:

$$\displaystyle \begin{bmatrix} \frac{T(1,1)}{1 \cdot 1^s}&+\frac{T(1,2)}{1 \cdot 2^s}&+\frac{T(1,3)}{1 \cdot 3^s}+&\cdots&+\frac{T(1,k)}{1 \cdot k^s} \\ \frac{T(2,1)}{2 \cdot 1^s}&+\frac{T(2,2)}{2 \cdot 2^s}&+\frac{T(2,3)}{2 \cdot 3^s}+&\cdots&+\frac{T(2,k)}{2 \cdot k^s} \\ \frac{T(3,1)}{3 \cdot 1^s}&+\frac{T(3,2)}{3 \cdot 2^s}&+\frac{T(3,3)}{3 \cdot 3^s}+&\cdots&+\frac{T(3,k)}{3 \cdot k^s} \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ \frac{T(n,1)}{n \cdot 1^s}&+\frac{T(n,2)}{n \cdot 2^s}&+\frac{T(n,3)}{n \cdot 3^s}+&\cdots&+\frac{T(n,k)}{n \cdot k^s} \end{bmatrix} = \begin{bmatrix} \frac{\zeta(s)}{1} \\ +\frac{\zeta(s)\sum\limits_{d|2} \frac{\mu(d)}{d^{(s-1)}}}{2} \\ +\frac{\zeta(s)\sum\limits_{d|3} \frac{\mu(d)}{d^{(s-1)}}}{3} \\ \vdots \\ +\frac{\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}}{n} \end{bmatrix}$$

On the right hand side we see that it sums to the right hand side of what we set out to prove, namely:

$$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$

Things that remain unclear are: What factor should the left hand side be multiplied with in order to have the same magnitude as the right hand side? And why in the Fourier transform does the first term of the von Mangoldt function appear to be $\log q$?

$$\Lambda(n) = \begin{cases} \log q & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

$$n=1,2,3,4,5,...q$$

As a heuristic, $$\Lambda(n) = \log q \;\;\;\; \text{if }n=1$$ probably has to do with that in the Fourier transform $q$ terms of $\Lambda$ are used and the first column in square matrix $T(1..q,1..q)$ sums to a Harmonic number.