= Solution
Write $\mathbf1(n)=1$ and let $\varepsilon$ be the <Identity for Dirichlet convolution>. The <Möbius function> uses the empty-product convention $\mu(1)=1$; the <Von Mangoldt function> has $\Lambda(1)=0$. We first establish all the <arithmetic function> identities needed below. Every <integer divisor> of $n$ contributes one to $\tau(n)$, so $\tau=\mathbf1*\mathbf1$. If $n>1$ has $r$ distinct <prime factors>, the squarefree divisors correspond to their subsets and
$$
\sum_{d\mid n}\mu(d)=\sum_{j=0}^r\binom rj(-1)^j=(1-1)^r=0.
$$
At $n=1$ the sum is one. Thus $\mu*\mathbf1=\varepsilon$. Writing $n=\prod_p p^{v_p(n)}$, the only divisors with nonzero <Von Mangoldt function> are $p,p^2,\ldots,p^{v_p(n)}$, and hence
$$
\sum_{d\mid n}\Lambda(d)=\sum_p v_p(n)\log p=\log n.
$$
The <Dirichlet convolution> is commutative by replacing $d$ with $n/d$, and associative because either bracketing sums $f(a)g(b)h(c)$ over $abc=n$. Consequently
$$
\boxed{\mu*\tau=\mathbf1,\qquad\mu*\log=\Lambda.}
$$
These are consequences of the definitions, rather than unproved invocations of <Möbius inversion>.
To identify the constant, integrate on each interval $[j,j+1)$:
$$
\int_j^{j+1}\frac{\{x\}}{x\lfloor x\rfloor}\,dx
=\frac1j-\log\frac{j+1}{j}.
$$
The tail is $O(1/J)$, since the integrand is at most $2/x^2$ for $x\ge2$. Summation therefore gives $\gamma=\lim_{J\to\infty}(H_J-\log(J+1))$, where $H_J$ is the <harmonic number>, and $H_J=\log J+\gamma+O(1/J)$. Thus this is <Euler's constant>.
Put $r=\lfloor\sqrt N\rfloor$. Counting pairs of <positive integers> whose product is at most $N$, and subtracting the square counted twice, gives the <Dirichlet hyperbola method> identity
$$
D(N):=\sum_{n\le N}\tau(n)
=2\sum_{a\le r}\left\lfloor\frac Na\right\rfloor-r^2
=2NH_r-r^2+O(r).
$$
Since $r=\sqrt N+O(1)$, the <harmonic number> estimate yields
$$
D(N)=N\log N+(2\gamma-1)N+O(\sqrt N).
$$
The increasing <natural logarithm> also satisfies $\sum_{n\le N}\log n=N\log N-N+O(\log(2N))$ by comparison with its integral. Subtracting proves
$$
\boxed{\Delta(N)=O(\sqrt N).}
$$
For the second implication, denote the <Mertens function> by $M(x)$ and the <Second Chebyshev function> by $\psi(x)$. Apply the proved <Dirichlet convolution> identities to $b(n)=\tau(n)-\log n-2\gamma\mathbf1(n)$:
$$
\mu*b=\mathbf1-\Lambda-2\gamma\varepsilon.
$$
Summing this identity over integers up to $N$ gives the exact relation
$$
\psi(N)=N-2\gamma-E(N),\qquad
E(N)=\sum_{d\le N}\mu(d)\Delta(\lfloor N/d\rfloor).
$$
Fix an integer $K\ge2$. The contribution to $E(N)$ from $d\le N/K$ has absolute value at most
$$
C\sqrt N\sum_{d\le N/K}d^{-1/2}\ll\frac N{\sqrt K}.
$$
For $d>N/K$, group by $j=\lfloor N/d\rfloor$. This gives exactly
$$
\sum_{j=1}^{K-1}\Delta(j)
\left(M(N/j)-M(N/(j+1))\right)=o_K(N)
$$
under the assumed cancellation $M(x)=o(x)$; extending that assumption from integers to real $x$ uses $M(x)=M(\lfloor x\rfloor)$. Therefore $\limsup_{N\to\infty}|E(N)|/N\ll K^{-1/2}$. Letting $K\to\infty$ proves the <prime number asymptotic from Mertens cancellation>:
$$
\boxed{\frac{\psi(N)}N=1+o(1).}
$$
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