No! This is very badly false -- the Riemann zeta function has non-trivial zeros. For example, suppose that $M(x) = \sum_{n=1}^{\infty} \mu(n) e^{-nx}$ tends to $-2$ as $x\to 0$ (I've rewritten your hypothesis with $x=1/s$). In particular, you're assuming that $M(x)$ is always bounded. But in that case note that
$$
\int_0^{\infty} M(x) x^{s}\frac{dx}{x} = \sum_{n=1}^{\infty}\mu(n) \int_0^{\infty} e^{-nx} x^{s} \frac{dx}{x} = \frac{\Gamma(s)}{\zeta(s)},
$$
where the integral a priori converges for Re$(s)>1$ (because as $x\to \infty$ clearly $M(x)\ll e^{-x}$ decreases exponentially, and trivially as $x\to 0^+$ we can use $|M(x)| \ll x^{-1}$). The assumption that $M(x)$ is bounded as $x\to 0^+$ now implies that the integral actually makes sense in Re$(s)>0$, or in other words that $\zeta(s)$ has no zeros! See also my answer to Is it possible to show that $\sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}$ diverges? .

Alternatively, one can write down an explicit formula for $M(x)$ in terms of zeros of $\zeta(s)$. Namely for some $c>1$
$$
M(x) = \frac{1}{2\pi i} \int_{c- i\infty}^{c+i\infty} \frac{1}{\zeta(s)} x^{-s} \Gamma(s) ds,
$$
and moving the line of integration to the left, we find
$$
M(x) = \sum_{\rho} \frac{\Gamma(\rho)}{\zeta^{\prime}(\rho)} x^{-\rho}+ \frac{1}{\zeta(0)} + O(x),
$$
where the sum is over non-trivial zeros of zeta (assumed to be simple for convenience). The second term above arises from the pole of the Gamma function at $s=0$, and note that it equals $-2$. The error term can be made explicit in terms of the poles at $-1$, $-2$, etc. This shows why $M(x)$ will have to be of size at least $x^{-1/2}$ occasionally, and further explains why the numerical evidence is misleading: the first zero of $\zeta$ has large ordinate (about $14.1\ldots$), and $\Gamma(1/2+14.1\ldots i)$ is very small in size.