Just an heuristic answer:

The joint eigenvalue distribution of a GOE random matrix is given by $$ \frac{1}{Z_n}\prod_{i<j}|x_i-x_j|\prod_{i=1}^ne^{-x_i^2/2}dx_i. $$ Thus, $E_{GOE}(|\det(y+A)|e^{-x(tr A)^2})$ equals to $$ \frac{1}{Z_n}\int\cdots\int\prod_{i=1}^n|y+x_i|e^{-x\sum_{i,j}x_ix_j} \prod_{i<j}|x_i-x_j|\prod_{i=1}^ne^{-x_i^2/2}dx_i, $$ and, if I introduce the measure $\mu_n=\frac{1}{n}\sum_{i=1}^n\delta_{x_i}$, to \begin{multline} \frac{1}{Z_n}\int\cdots\int\exp\left\{-n^2\left[-\frac{1}{n}\int\log|y+u|\mu_n(du)+x\iint uv\mu_n(du)\mu_n(dv)\\ -\iint_{u<v}\log|u-v|\mu_n(du)\mu_n(dv)+\frac{1}{2n}\int u^2\mu_n(du)\right]\right\}\prod_{i=1}^ndx_i. \end{multline} Similarly, $$ Z_n=\int\cdots\int\exp\left\{-n^2\left[ -\iint_{u<v}\log|u-v|\mu_n(du)\mu_n(dv)+\frac{1}{2n}\int u^2\mu_n(du)\right]\right\}\prod_{i=1}^ndx_i. $$

Now, let me denote by $n^{\alpha}_*\mu$ the push-forwards by $x\mapsto n^{\alpha}x$ of a probability measure $\mu$. It is known that $n^{-1/2}_*\mu_n$ converges weakly towards the semi-circle distribution $\mu_{SC}$ with probability one.

As a consequence, heuristically I would say that as $n\rightarrow\infty$ (which may be made rigorous by large deviation estimates)

\begin{multline} \frac{1}{n^2}\log E_{GOE}(|\det(|y+A|e^{-x(tr A)^2})\\ \simeq \frac{1}{n}\int\log|y+u|\,(n^{1/2}_*\mu_{SC})(du)-x\iint uv \,(n^{1/2}_*\mu_{SC})(du)(n^{1/2}_*\mu_{SC})(dv), \end{multline} the latter being computable explicitly.