If $$t^p-(r^p+s^p) \equiv 0 \;( \text{mod}\; p) \tag{1}$$ where $r$, $s$ and $t$ are all relatively prime non-zero positive integers and $p$ is any odd prime.
then one solution super set, that includes at least some of the relatively prime solutions is $$t-(r+s)=pq \tag{2}$$, where $q$ is a positive integer.
I think the most straight forward way to prove this is to assume (2) correct and write
$$t^p-(t-pq)^p+(r+s)^p-(r^p+s^p)$$
and to then separately confirm that $$t^p-(t-pq)^p \equiv 0 \;( \text{mod}\; p) \tag{3}$$
$$(r+s)^p-(r^p+s^p) \equiv 0 \;( \text{mod}\; p) \tag{4}$$
(3) being an obvious result and (4) being a well known result.
The question is are there other relatively prime solutions to (1) outside of those encompassed by (2)?
No. The most simple way to see this is to work in the field $\mathbf Z/p\mathbf Z$.
Indeed, in this field, the Frobenius map: \begin{align} \mathbf Z/p\mathbf Z&\longrightarrow\mathbf Z/p\mathbf Z\\ x&\longmapsto x^p \end{align} is a ring homomorphism, i.e. for any $x,y\in\mathbf Z/p\mathbf Z$, $\;(x+y)^p\equiv x^p+y^p$ (this results from the binomial formula and the fact that, if $p$ is prime, $\;\binom pk\equiv 0\bmod p$ for all $1\le k\le p-1$) and $(xy)^p\equiv x^py^p$.
Therefor, one has $\;t^p-(r^p+s^p)\equiv \bigl(t-(r+s)\bigr)^p$, and as we're in a field, $$\bigl(t-(r+s)\bigr)^p\equiv 0\iff t-(r+s)\equiv 0\pmod p$$
Note: this has nothing to do with $r,s,t$ being relatively prime.