Not an answer, but a possible start to a character computation @LSpice, too long for a series of comments.
Fix $g$ in $G$, and let $a_i$ and $b_i$ be the eigenvalues of $\rho_i(g)$ for $i = 1, 2$. Then the eigenvalues of ${\rm ad} \rho_i(g)$ are $1$, $a_i b_i^{-1}$ and $b_i a_i^{-1}$. By assumption the sets $\{a_1 b_1^{-1}, b_1 a_1^{-1}\}$ and $\{a_2 b_2^{-1}, b_2 a_2^{-1}\}$ are the same. If $a_i b_i^{-1} \neq \pm 1$ then these two sets consist of distinct elements, and up to switching $a_2$ and $b_2$, we can identify that $a_1 b_1^{-1} = a_2 b_2^{-1}$. Therefore, if such a character $\eta$ were to exists, we would have to have $\eta(g) = a_1 a_2^{-1} = b_1 b_2^{-1}$. This is still true if $a_i b_i^{-1} = 1$: now $a_i = b_i$, so we do not have to worry about picking which of the two eigenvalues of $\rho_2(g)$ to call $a_2$, as they are the same. It's only if $a_i b_i^{-1} = -1$ (equivalently, ${\rm tr} \rho_i(g) = 0$) that we do not know which eigenvalue to call $a_2$. But since $b_2 = - a_2$, the quantity $\pm a_2$ is well-defined; in this case $\eta(g) = \pm a_1 a_2^{-1}$ is only determined up to sign.
So we have now defined a quasi-function $\eta: G\ "\!\to\!"\ K^\times$ with $\eta(g) = a_1 a_2^{-1}$ whenever ${\rm tr} \rho_i (g) \neq 0$ and $\eta(g)$ defined only up to sign otherwise. (I am just using "quasi-function" informally here.) Note that this quasi-function satisfies ${\rm tr} \rho_1 = \eta \cdot {\rm tr} \rho_2$, so that if one could prove that the quasi-function can be refined to a function that is a character $G \to K^\times$, then by Brauer-Nesbitt --- assuming we are free to assume $\rho_i$ are semisimple --- we would be done.
Moreover, note that $\eta^2: G \to K^\times$ is not only a perfectly well-defined function but also a character: indeed, $\eta^2 = {\rm det} \rho_1 \otimes {\rm \det \rho_2}^{-1}$.
To be continued, as I must dash, but possibly someone already has an idea for how to take it from here?
Post-dash sequel:
[At this point, by the way, we have a proof for $K$ of characteristic $2$: there is no sign ambiguity, so that $\eta: G \to K^\times$ is a proper function; and since $\eta^2$ is a character, so is $\eta$. Finally since $\rho_1$ and $\rho_2 \otimes \eta$ have the same trace and determinant, and we are assuming $\rho_i$ semisimple, Brauer-Nesbitt applies to conclude that they are isomorphic representations.]
Some ideas about how to proceed from here, to see that $\eta$ is multiplicative where defined:
From the definition, it follows that $\eta(1) = 1$.
The identity ${\rm tr} \rho_i(g^{-1}) = \det \rho_i(g)^{-1} {\rm tr} \rho_i(g)$ combined with the trace and determinant equations gives us ${\rm tr} \rho_2(g) \eta(g^{-1}) = {\rm tr} \rho_2(g) \eta(g)^{-1}$, so that $\eta(g^{-1}) = \eta(g)^{-1}$ so long as ${\rm tr} \rho_i(g) \neq 0$.
Since $\det \rho_i(g) = \frac{1}{2}\big(({\rm tr} \rho_i(g))^2 - {\rm tr} \rho_i(g^2)\big)$ and $\det \rho_1(g) = \det \rho_2(g) \eta(g)$ and ${\rm tr} \rho_1(g) = {\rm tr} \rho_2(g) \eta(g)$, we can conclude that ${\rm tr} \rho_2(g^2) \eta(g)^2 = {\rm tr} \rho_2(g^2) \eta(g^2)$ for all $g \in G$. Therefore, if ${\rm tr} \rho_i(g^2) \neq 0$, then $\eta(g)^2 = \eta(g^2)$. (I don't know how useful this is.)
The general version of the two identities above is the pseudocharacter identity $${\rm tr} \rho_i(gh) + \det \rho_i(g) {\rm tr} \rho_i(g^{-1} h) = {\rm tr} \rho_i(g) {\rm tr} \rho_i(h),$$ which holds for any $g, h$ in $G$. Whence, for $g, h \in G$, $${\rm tr} \rho_2(gh) \big(\eta(gh) - \eta(g) \eta(h)\big) = \det \rho_2(g) {\rm tr} \rho_2(g^{-1} h) \big(\eta(g) \eta(h) - \eta(g)^2 \eta(g^{-1} h) \big).$$
Can anything useful come out of this?
