I think that the Goeritz invariant of link $k(L1 \# L2)$ is $(k(L1), k(L2))$.
Do you have an example of non trivial knot that have determinant 1?
If there is no such example, then the following holds: if a depth of Goeritz invariant > 1, then the link is prime.
As Lukas mentioned in his comment, there are several such examples.
My favourite are torus knots $T(p,q)$ with $p,q$ coprime and both odd have determinant 1.
The easy way to see this is the following: $\det K = \Delta_K(-1)$, where $\Delta$ is the Alexander polynomial. Since $\Delta_{T(p,q)}(t) = \frac{(t^{pq}-1)(t-1)}{(t^p-1)(t^q-1)}$, we have $$ \det T(p,q) = \Delta_{T(p,q)}(-1) = \frac{((-1)^{pq}-1)((-1)-1)}{((-1)^p-1)((-1)^q-1)} = \frac44 = 1. $$
Another way is to think of $\det K$ as the order of $H_1(\Sigma(K))$, where $\Sigma$ is the double cover of $S^3$ branched over $K$. For a torus knot $\Sigma(T(p,q))$ is the link $\Sigma(2,p,q)$ of the singularity of $x^2+y^p+z^q$, and Brieskorn showed that $\Sigma(p,q,r)$ is an integer homology sphere when $\gcd(p,q) = \gcd(q,r) = \gcd(r,p) = 1$, which is exactly the case we're in.