Not an answer, but the formulae are too bulky for a comment.
After a lot of experimenting I found a formula for the coefficients of $\phi_{n}(x^{-1})$$\phi_{n}(y)$. (Currently without formal proof, but the evidence is really strong.)
I conjecture that $\phi_{n}(x^{-1})$$\phi_{n}(y)$ has the following form $$ \phi_{n}(x^{-1})= (-1)^{n} + \sum_{k=2}^{2 n} c_{n,k} x^{k}, $$$$ \phi_{n}(y)= (-1)^{n} + \sum_{k=2}^{2 n} c_{n,k} y^{k}, $$ where $$ c_{n,k}=c^{<}_{n,k}:=\frac{(-1)^{n+1} n!}{(n-k+1)!}\ {}_{2}F_{3}(1-\frac{k}{2},\frac{3}{2}-\frac{k}{2};2,2-k,n-k+2;4) $$ for $2\leq k \leq n $ and $$ c_{n,k}=c^{>}_{n,k}:=\frac{(-1)^{k} n!}{(k-n)!}{{n-1}\choose{k-n-1}} {}_{2}F_{3}(\frac{k}{2}-n,\frac{k}{2}-n+\frac{1}{2};1-n,k-n,k-n+1;4) $$ for $n< k \leq 2 n $. The ${}_{2}F_{3}$ are generalized hypergeometric functions.
The series of the hypergeometric functions in the formula for $c^{<}_{n,k}$ terminates, since one of the nominatorial parameters of the hypergeometric is always a non-positive integer.
The sums of the hypergeometric functions in the formulae for $c^{<}_{n,k}$ and $c^{>}_{n,k}$ terminate. The non-positive denominatorial parameters, are always overruled by one of the nominatorial parameters, since, e.g. $1-n<\frac{k}{2}-n$ for the relevant values. In cases of equality of the nominatorial and denominatorial parameters they cancel and the function collapses to a well defined ${}_{1}F_{2}$. This overruling and cancellation has to be carefully implemented in math algebra systems like Mathematica, since it is not easily recognized by them.
The sign of $c_{n,k}$ changes depending on $n$ and $k$ in irregular manner as already mentioned in other answers. The oscillations appear for $k$ in a relative small range above and below $n$ starting for $n$ above 7.
The formulae were found by using $f$ with a decoration of the variable $x$ $$ f(x)=\exp( - x \ z - \frac{y}{x} ) $$ and expand the result for $\phi$ in $x$, $z$, and $y$ using Mathematica. The formulae for the coefficients of the resulting multinomials were easily guessed and after setting $z=1$ and $y=1$ only Mathematica solvable sums remained.