Suppose $S \in \mathcal{L}(\textsf{V})$ and $S$ is injective. Define $$\langle u,v\rangle_1 = \langle Su,Sv\rangle$$ Show that $\langle u,v\rangle_1$ is an inner product on $\textsf{V}$. The proof that I came across feels circular. Here I just focus on the property $ \langle u+v,w\rangle_1 = \langle u,w\rangle_1+\langle v,w\rangle_1$ :
$$\begin{matrix} \langle u+v,w\rangle_1=\langle S(u+v),Sw\rangle & & & (1) \\ \langle S(u+v),Sw\rangle=\langle Su+Sv,Sw\rangle & & & (2) \\ \langle Su+Sv,Sw\rangle=\langle Su,Sw\rangle + \langle Sv,Sw\rangle & & & (3) \\ \langle Su,Sw\rangle + \langle Sv,Sw\rangle=\langle u,w\rangle _1+ \langle v,w\rangle_1 & & & (4) \end{matrix}$$
But doesn't line $(3)$ use the very fact that we are trying to prove?
No, because it seems that you already assume that $(V,\langle \cdot, \cdot\rangle)$ is an inner product space. Otherwise the question doesn‘t make sense.
Put differently: You already know that $\langle \cdot, \cdot\rangle$ is an inner product (which you use in line 3); and you want to show that $\langle \cdot, \cdot\rangle_1$ also is an inner product on $V$.