For bounded potentials $W_j$, this is quite easy: the general asymptotics of the eigenvalues are $|\lambda_n(W) + n^2\pi^2/T^2|\le \|W\|_{\infty}$. This follows by just comparing them with those of $W=0$.

Thus $$ \log \prod \frac{\lambda-\lambda_n(W_1)}{\lambda-\lambda_n(W_2)} = \sum \log \frac{\lambda-\lambda_n(W_1)}{\lambda-\lambda_n(W_2)} = \sum O(1/(\lambda+n^2)) $$ indeed goes to $0$, by dominated convergence.

For more general $W$, this would require more precise asymptotic formulae.

As I already mentioned in a comment, the product "defining" $\det(D^2-W)$ is badly divergent, but we can consider the ratio, interpreted as $\prod\frac{\lambda-\lambda_n(W_1)}{\lambda-\lambda_n(W_2)}$.

Then this is quite easy for bounded potentials $W_j$: the general asymptotics of the eigenvalues are $|\lambda_n(W) + n^2\pi^2/T^2|\le \|W\|_{\infty}$. This follows by just comparing them with those of $W=0$.

Thus $$ \log \prod \frac{\lambda-\lambda_n(W_1)}{\lambda-\lambda_n(W_2)} = \sum \log \frac{\lambda-\lambda_n(W_1)}{\lambda-\lambda_n(W_2)} = \sum O(1/(\lambda+n^2)) $$ indeed goes to $0$, by dominated convergence.

For more general $W$, this would require more precise asymptotic formulae.