Let $P_0,P_1,...,P_n$ be $n+1$ points of $\mathbb{E}^n$ in "general positions" i.e they don't lie in any hyperplane of $\mathbb{E}^n$. Two Isometries that are equal at $P_i$'s are the same.
It suffices to show an Isometry of $\mathbb{E}^n$ that fixes $P_0,P_1,...,P_n$ is the identity $I$. Let $T$ be one such candidate. Define $S:\mathbb{E}^n \to \mathbb{E}^n$ as $S(x)=x-P_0$ which is a Translation. I observed that $S\circ T \circ S^{-1}$ is an Isometry with the formula $$STS^{-1}(x) = T(x+P_0) - P_0 \Rightarrow STS^{-1}(0) = 0 \ \text{and} \ STS^{-1}(P_i-P_0) = P_i-P_0.$$
We have shown that $S \circ T \circ S^{-1}$ is an Isometry fixing origin and the points $\{P_i-P_0: 1 \le i \le n\}$.
If somehow we are able to show that the set $\{P_i-P_0: 1 \le i \le n \}$ is a basis of $\mathbb{E}^n.$ We apply the above theorem to obtain $S \circ T \circ S^{-1} = I \Rightarrow T = I$.
Remains to show $\{P_i-P_0: 1 \le i \le n \}$ is a basis of $\mathbb{E}^n$.
Considering the linear relation $\sum\limits_{i=1}^n c_i(P_i-P_0) = 0$.
If for some $i_0$ we have $c_{i_0} \ne 0$ then $P_{i_0}-P_0 \in L\{P_i-P_0: i \ne i_0\}$ -----(1).
If I can show that $P_i-P_0$'s do not lie in any hyperplane I will obtain a contradiction to (1). How can I show this?
You use the word hyperplane in two different senses. First, for an affine hyperplane, then, at the end, for a linear hyperplane. In any case, an affine hyperplane is the set of the form $P_0 + \mbox{linear hyperplane},$ so your assumption gives you all you need.