Your definition must include that $f(1)\not= 0$, because $f^{-1}(n)$ is recursively defined in terms of $1/f(1)$. (The formula can be found on the Wikipedia.) So, denote by $\mathcal{A}$ the group of all arithmetic functions with $f(1)\not= 0$.
Let $\mathcal{U}$ be the subgroup of all $f\in \mathcal{A}$ such that $f(1)=1$ and $\mathcal{C}$ be the subgroup of scalar functions $\epsilon_c(n):=c\epsilon(n)$, where $\epsilon=\text{id}_\mathcal{A}$. Note that $\mathcal{C}\cap \mathcal{U}=\{\epsilon\}$ Define $(\frac{1}{c}f)(n):=\frac{f(n)}{c}$. If we consider an arbitrary $g\in \mathcal{A}$, then $\frac{1}{f(1)}f\in\mathcal{U}$. We have $\left(\frac{1}{g(1)}g\star \epsilon_{g(1)}\right)(n)=\sum_{ab=n}\frac{1}{f(1)}f(a)f(1)\epsilon(b)=(f\star \epsilon)(n)=f(n).$ Thus we can uniquely factor $f=\frac{1}{f(1)}f\star \epsilon_f(1)$, whence $\mathcal{A}=\mathcal{U}\oplus \mathcal{C}$.
Now define $\mathcal{K}\subseteq \mathcal{U}$ to be the subset of (not necessarily multiplicative) functions which vanish on prime powers. Since $f^{-1}(n)$ is defined in terms of $f(n)$, and since $\mathcal{K}$ is obviously closed under $\star$, it follows immediately that $\mathcal{K}$ is a subgroup of $\mathcal{U}$. Again $\mathcal{H}\cap \mathcal{K}=\{\epsilon\}$, where $\mathcal{H}$, as given by OP, is the subgroup of multiplicative functions. For $f\in \mathcal{U}$, define $\Pi_f$ so that $\Pi_f(p_1^{e_1}\cdots p_s^{e_s})=\prod_{k=1}^r f(p_k^{e_k}).$ Obviously $\Pi_f\in \mathcal{H}$. We have that $\begin{eqnarray*} \left(\Pi_f^{-1}\star f \right)(p^{e})&=&\sum_{k=0}^e\Pi_f(p^k)^{-1}f(p^{e-k})\\&=&\sum_{k=0}^e\Pi_f^{-1}(p^k)\Pi_f(p^{e-k})\\&=&(\Pi_f^{-1}\star\Pi_f)(p^e)\\&=&\epsilon(p^e)\\&=&0.\end{eqnarray*}$ It follows that $\Pi_f^{-1}\star f\in \mathcal{K}$. Now of course we can write $f=\Pi_f\star\left(\Pi_f^{-1}\star f\right)$, so $f$ uniquely factors into $\mathcal{H}$ and $\mathcal{K}$ parts, and we have that $\mathcal{U}=\mathcal{H}\oplus \mathcal{K}$.
So we have obtained that $\mathcal{A}=\mathcal{H}\oplus \mathcal{K}\oplus \mathcal{C}$.