No series in $y$ can yield $x$. To see this, first note that, replacing $y(x)$ by $y(x)-a_0$, one can assume without loss of generality that $a_0=0$. Introduce the series $A(u)=\sum\limits_{n=0}^{+\infty}a_nu^n$. Thus, one looks for an expression of $x$ as a function of $y(x)=\log(x)+A(\frac1x)$.
A solution is to note that $\mathrm e^{y(x)}=x\mathrm e^{A(1/x)}$, hence $ x=\mathrm e^{y(x)}B(\mathrm e^{-y(x)}), $ where the series $B(v)=\sum\limits_{n=0}^{+\infty}b_nv^n$ solves the relation $ \mathrm e^{-A(u)}=\sum\limits_{n=0}^{+\infty}b_nu^n\mathrm e^{-nA(u)}. $ Finally, $b_0=1$ hence $ \color{red}{x=\mathrm e^{y(x)}+\sum\limits_{n=0}^{+\infty}b_{n+1}\mathrm e^{-ny(x)}}. $ Note 1: The series $B$ solves the relation $C(u)=uB(C(u))$ with $C(u)=u\mathrm e^{-A(u)}$.
Note 2: Using the inverse $A^{-1}$ of $A$, one can characterize the series $B$ through its inverse $B^{-1}$ by the relation $ B^{-1}(w)=wA^{-1}(-\log w). $ Note 3: To compute the polynomial $B_N(v)=\sum\limits_{n=0}^{N}b_nv^n$, one can use the polynomial $A_N(u)=\sum\limits_{n=0}^{N}a_nu^n$ and solve the finite relation $ \mathrm e^{-A_N(u)}=\sum\limits_{n=0}^{N}b_nu^n\mathrm e^{-nA_N(u)}+o(u^N), $ that is, $ \sum\limits_{k=0}^N\frac{(-1)^k}{k!}\left(\sum\limits_{i=0}^{N}a_iu^i\right)^k=\sum\limits_{k=0}^N\frac{(-1)^k}{k!}\left(\sum\limits_{i=0}^{N}a_iu^i\right)^k\sum\limits_{n=0}^{N}b_nn^ku^n+o(u^N). $ For example, $b_1=-a_1$, $b_2=-a_2-\frac12a_1^2$.
Note 4: Assume that $a_n=\frac1n$ for every $n\geqslant1$, then $A(u)=-\log(1-u)$ and $B(v)=\frac12+\sqrt{\frac14-v}$, hence $x=\frac12\mathrm e^{y(x)}\left(1+\sqrt{1-4\mathrm e^{-y(x)}}\right)$.
Note 5: This is to answer a question asked by the OP in a comment. Fix $x$ and let $z=\mathrm e^{-y(x)}$. Then $x^{-1}\mathrm e^{-A(1/x)}=z$ and $x=z^{-1}B(z)$, hence $\mathrm e^{-A(1/x)}=xz=B(z)=B(x^{-1}\mathrm e^{-A(1/x)})$. Thus, for every $u$, $\mathrm e^{-A(u)}=B(u\mathrm e^{-A(u)})$.