I apologize for the delay in writing this up. Here is a proof of the remaining inequality (the one not proved in your answer, ireallydonknow). As I mentioned in a comment, this proof is a simple application of ideas found on page 7 of Hardy & M. Riesz's The General Theory of Dirichlet's Series. Here are a few remarks (mostly paraphrasing that book) to clarify the import of these formulas for other people who stumble across this question and wonder about its motivation. Basically, this formula is an analogue of the root test for the convergence of a power series (in fact, in the case $\lambda_n = n$ the two tests coincide). The situation is much more subtle in the context of Dirichlet series, however—a power series converges absolutely or diverges except possibly on a subset of a single circle, where it may converge conditionally (I think though that that subset can be quite complicated); a Dirichlet series, on the other hand, may converge conditionally (and not absolutely) on a strip or even on the entire plane. Hardy and Riesz, for instance, give as an example on page 9 the series $\Sigma (-1)^n (\log n)^{-s}/\sqrt{n}$.
With that, let me turn to the proof I mentioned in the comments. Right off the bat let me say that your formula is slightly off as written, in that the $\lambda_n$ appearing in the definition of $\eta$ should really be $\lambda_{n+1}$. To be completely precise, assume that $\sigma_c\leq 0$ and put
\begin{align*}
R_n = \sum_{k = n+1}^\infty a_k, \quad \eta = \limsup_{n\to\infty} {\log |R_n|\over \lambda_{n+1}};\tag{1}
\end{align*}
then $\eta = \sigma_c$. You'll see that the proof given in your own answer, ireallydonknow, carries through perfectly well with $\eta$ defined as in $(1)$, and shows that under these circumstances $\sigma_c\geq \eta$. The purpose of this answer, then, is to prove the reverse inequality $\sigma_c\leq \eta$. This reverse inequality will be proved if it can be shown that the Dirichlet series $\Sigma a_k e^{-\lambda_k s}$ converges whenever $s$ is a real number strictly larger than $\eta$, and that is what we'll show.
Thus, suppose $s>\eta$; we may assume moreover that $s<0$, since by hypothesis $\sigma_c\leq 0$. A tautological implication of the definition $(1)$ of $\eta$ is that $\log{|R_n|} \leq \lambda_{n+1}(\eta + o(1))$ as $n\to\infty$, and so $|R_n| \leq e^{\lambda_{n+1}(\eta + o(1))}$. Now use summation by parts to write (I'm denoting by $A$ the sum $\Sigma a_k$)
\begin{align*}
\sum_{k=1}^n a_k e^{-\lambda_k s} &= \sum_{k = 1}^n (R_{k-1} - R_k) e^{-\lambda_k s} \\
&= Ae^{-\lambda_1 s} - R_n e^{-\lambda_n s} + \sum_{k=2}^{n-1} R_k \left( e^{-\lambda_{k+1} s} - e^{-\lambda_k s}\right).
\end{align*}
First, $R_n e^{-\lambda_n s} \to 0$ as $n\to \infty$:
\begin{align*}
\log{|R_n|} - \lambda_n s < \log{|R_n|} - \lambda_{n+1} s \leq \lambda_{n+1}(\eta + o(1) - s) \to -\infty \quad \text{as $n\to\infty$,}
\end{align*}
using $\eta - s<0$ and $\lambda_{n+1}\to\infty$ for the limit relation at the end. The first inequality relies on $s<0$ and $\lambda_n<\lambda_{n+1}$ to conclude $-\lambda_n s<-\lambda_{n+1} s$.
We're now reduced to showing that $\Sigma R_k (e^{-\lambda_{k+1} s} - e^{-\lambda_k s})$ converges. (This is the part where we actually need my definition $(1)$ of $\eta$ instead of yours.) The sum actually converges absolutely. To see this, we follow the example of Hardy and Riesz and bound the general term as follows:
\begin{align*}
|R_k| (e^{-\lambda_{k+1} s} - e^{-\lambda_k s}) &\leq e^{\lambda_{k+1}(\eta + o(1))} (e^{-\lambda_{k+1} s} - e^{-\lambda_k s}) \\
& \leq |s| (\lambda_{k+1} - \lambda_k) e^{\lambda_{k+1}(\eta - s + o(1))}\\
&\leq |s| \int_{\lambda_k}^{\lambda_{k+1}} e^{x(\eta - s + o(1))}\,dx
\end{align*}
provided $k$ is large enough that $\eta - s + o(1) < 0$. Note that we use my definition $(1)$ of $\eta$ to get the third inequality, because we rely on $\lambda_{k+1}$ in the exponential factor arising in the bound on $|R_k|$; the definition in the question would lead to $\lambda_k$ instead, which isn't enough to make the third inequality hold. Anyway, the integrals are summable, and the proof is finished.
I suppose the meaning of singularity here is not, that $f(1)=\infty$ but rather just that f is not holomorphic at $1$. Naively differentiating $f$ leads to
$$f'(s)=-\sum_{n=2}^\infty \frac{1}{n^s\log(n)},$$
which doesn't converge at $1$ anymore.
Edit: Actually for $Re(s)>1$ the derivative definitely has the above form. Therefore $f'(1+t)\to \infty$ for $t\to 0$ so $f$ cannot be continuously differentiable at $1$, therefore it is also not holomorphic.
Best Answer
A well known theorem of Landau shows that a Dirichlet series with positive coefficients will have a pole at $s = \sigma_0$, where $\mathrm{Re} s = \sigma_0$ is the abscissa of convergence. It is clear from your expression that the first pole of this series is at $s = 1$, and thus the abscissa of convergence is $1$.
A different perspective is to consider what the Dirichlet series represents. The average order of $\lvert \mu(n) \rvert$ is $6/\pi^2$. This is because $$\sum_{n \leq X} \mu(n) $$ counts the number of of squarefree integers up to $X$. Partial summation then implies that the Dirichlet series can't converge if $\mathrm{Re} s < 1$. Conversely, it's straightforward to show that the series converges for $\mathrm{Re} s > 1$ by comparison with the Riemann zeta function.