Let $f$ be a multiplicative arithmetic function such that $\sum_{n = 1}^{\infty} |f(n)|$ converges.
Given some $\epsilon > 0$, can we necessarily choose some $\delta$ such that for all arithmetic functions $g$ that obey the same above criteria as $f$, then:
$$\bar{\rho}(f, g) < \delta \ \Rightarrow \ \bar{\rho}(f^{-1}, g^{-1}) < \epsilon$$
where $h^{-1}$ is the Dirichlet inverse of $h$, and $\rho(f, g) = \sup \{ \min(|f(n) - g(n), 1) \ | \ n \in \mathbb{N}\}$ (it is the uniform metric).
Try with $$\sum_{n\ge 1} f(n)n^{-s}=\sum_{n\ge 1} f^{-1}(n)n^{-s}=1,\qquad \sum_{n\ge 1} g_r(n)n^{-s}=1-\delta \sum_{n\ge 1} r^n n^{-s}$$ with $\delta>0$ and $1-r>0$ small.
$$\frac1{G_r(s)}=1+\sum_{k\ge 1}\delta^k (\sum_{n\ge 1} r^n n^{-s})^k$$ Since everything is non-negative we get that $$g_r^{-1}(n)\ge \delta^2\sum_{d|n} r^d r^{n/d} \qquad \implies \sup_{r<1}\sup_n |f^{-1}(n)-g_r^{-1}(n)|=\infty$$
If you assume additionally that $\sum_n |f^{-1}(n)|=C$ converges and $\sum_n |g(n)-f(n)|<\delta<1/C$ then $\sup_{\Re(s)\ge 0} |F(s)|\ge 1/C$ so that $$\sum_{n\ge 1} (g^{-1}(n)-f^{-1}(n))n^{-s}=\frac1{F(s)-(F(s)-G(s))}-\frac1{F(s)}=\sum_{k\ge 1} \frac{(F(s)-G(s))^k}{F(s)^{k+1}}$$ This implies that $$\sum_{n\ge 1} |g^{-1}(n)-f^{-1}(n)| \le \sum_{k\ge 1} (\sum_n |f(n)-g(n)|)^k (\sum_n |f^{-1}(n)|)^{k+1}\le \sum_{k\ge 1} \delta^k C^{k+1}$$