You can have inverses with respect to convolution. Here are my ideas for constructing an inverse to an arithmetic function $f$, with $f(1) \neq 0$. Suppose that $g*f*g=g$, or

(1)
\begin{align} g(n)= \sum_{d|n,d>0} g(\frac{n}{d}) [ (f*g)(d) ] = \sum_{d|n,d>0} g(\frac{n}{d}) \left ( \sum_{d_1|d,d_1>0} f(\frac{d}{d_1})g(d_1) \right ) \end{align}

This $g$ actually works as the inverse of $f$. First of all, note that $g(1)=g(1)f(1)g(1)$, or $f(1)g(1)=1$. This means $(f*g)(1)=f(1)g(1)=1=i(1)$, where $i$ is the identity function. To finish showing that $g$ is the inverse of $f$, you just need to check that $(f*g)(n)=0=i(n)$ for $n>1$.

This seems like a nice job for induction. First, the base case. If $n=2$, then

(2)
\begin{align} g(2)=\sum_{d|2,d>0} g(\frac{2}{d}) [ (f*g)(d) ] = g(2) + g(1) [f(1)g(2)+f(2)g(1)] \end{align}

This implies that

(3)
\begin{equation} 0=g(1) [f(1)g(2)+f(2)g(1)] \end{equation}

where since $g(1)f(1)=1$ we know that $g(1) \neq 0$, so we really have

(4)
\begin{align} 0=f(1)g(2)+f(2)g(1)= \sum_{d|2,d>0}f(\frac{2}{d})g(d) = (f*g)(2) \end{align}

The induction step falls right out from that procedure. Assuming that $(f*g)(m)=0$ for $1<m<n$ we have

(5)
\begin{align} g(n)=\sum_{d|n,d>0} g(\frac{n}{d}) [ (f*g)(d) ] = g(n) + \sum_{d|n,d>1} g(\frac{n}{d}) [ (f*g)(d) ] \end{align}

Hence

(6)
\begin{align} 0=\sum_{d|n,d>1} g(\frac{n}{d}) [ (f*g)(d) ] \end{align}

And by assumption, the induction hypothesis, $(f*g)(d)=0$ for $1<d<n$ so we have

(7)
\begin{equation} 0=g(1)[(f*g)(n)] \end{equation}

Which implies $(f*g)(n)=0$, again by the cancellation property. So by induction $(f*g)(n)=0$ for $n>1$ and we

have $(f*g)(m)=i(m)$ for all $m$. Therefore $f*g=i$ and $g$ is the inverse of $f$. So yeah, the set of all arithmetic funtions $f$ with $f(1) \neq 0$ is a group with convolution as the operation.