The following is a step-by-step proof/derivation showing the difference in abscissae of convergence $a_c$ and absolute convergence $a_a$ is never more then 1.
Question: Is this simple proof correct?
It looks "too simple" and I suspect I have made an error.
Step 1:
Assume a Dirichlet series converges at complex $s_0$. For it to converge absolutely, the same series with each term replaced by its magnitude must also converge.
Step 2:
Let's consider the terms of the series at a point $s=s_{0}+d$ where $\sigma_{d}$ is the real part of $d$.
$$\left|\frac{a_{n}}{n^{s}}\right|=\left|\frac{a_{n}}{n^{s_{0}+d}}\right|=\left|\frac{a_{n}}{n^{s_{0}}}\right|\cdot\frac{1}{n^{\sigma_{d}}}$$
Step 3:
We know that for a series convergent at $s_{0}$ the terms $a_{n}/n^{s_{0}}$ are bounded. For a series absolutely convergent at $s$, the terms $\left|a_{n}/n^{s}\right|$ must be bounded.
Step 4:
So, for the series $\sum\left|a_{n}/n^{s}\right|$ to converge, so must $\sum1/n^{\sigma_{d}}$, and we know it does for $\sigma_{d}>1$. Therefore the distance between $\sigma_{a}$ and $\sigma_{c}$ must be no more than 1.
$$0\leq\sigma_{c}-\sigma_{a}\leq1$$
Update - I know Apostol's IANT has a proof but don't understand it properly, which is why I've written the above as my understanding.
Update 2 - I think my error is at step 3.
The actual proof is very similar to that of OP's.
By the definition of $\sigma_c$ and convergence, we see that there exists $c_\delta,M>0$ such that $|a_n|\le c_\delta n^{\sigma_c+\delta}$ for all $n>M$. As a result if we put $s=\sigma_c+1+2\delta$ then we have
$$ \sum_{n=1}^\infty{|a_n|\over n^s}\le\sum_{n=1}^M{|a_n|\over n^s}+c_\delta\sum_{n=M+1}^\infty{1\over n^{1+\delta}} $$
This means that for all $\delta>0$ we have $\sigma_a\le\sigma_c+1+2\delta$. Consequently by the characteristics of real numbers we have $\sigma_a-\sigma_c\le1$.