Your reasoning is a little faulty since you are assuming that $\phi ( \frac{1}{3} ) = \frac 13$ (since a priori there is not reason to think that $\frac 13$ must be a fixed point of such an isomorphism). But even if $\phi ( \frac 13 ) = \frac 13$, then this would tell us that $$\begin{align}
\phi ( 3 )
&= \phi (\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13) \\
&= \phi (\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi (\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi(\tfrac 13) \\
&= \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \\
&= 3^{-9}
\end{align}$$
and therefore $\phi ( -3 ) = ( \phi ( 3 ) )^{-1} = ( 3^{-9} )^{-1} = 3^9$.
If you recall the following rule of exponentiation: $$a^{x+y} = a^x \cdot a^y$$ you should begin to think that a mapping of the form $x \mapsto a^x$ looks "homomorphism-ish," and it is not too difficult to show that if $a > 0$, then such a mapping is an isomorphism between $( \mathbb{R} , + )$ and $( P , \cdot )$.