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.