A linear operator $L$ between complex spaces with inner product $U$ and $V$ is an isometry, only if $\left < Lu_i, Lu_j \right > = \left < u_i, u_j \right >$ for all $u_j, u_i$ from a basis of $U$ (not necessary orthonormal).
I would like to see a proof of this statement.
EDIT: ONLY IF direction proof without directly referring the polarization identity is as follows.
First we know that $$ \|Lu_m\|^2_V = \langle L u_m ,L u_m\rangle = \|u_m\|^2_U = \langle u_m , u_m\rangle\tag{1} $$ because of isometry. Now there is: $$ \|L(u_m-u_n)\|^2_V = \langle L(u_m-u_n) , L(u_m-u_n) \rangle = \langle u_m-u_n , u_m-u_n \rangle = \|u_m-u_n\|_U^2.\tag{2} $$ Expanding and using (1) lead to $$ \langle L u_m, Lu_n \rangle + \langle L u_n, Lu_m\rangle = \langle u_m , u_n \rangle + \langle u_n , u_m \rangle\tag{3} $$ Change (2) a little bit: $$ \|L(u_m-iu_n)\|^2_V = \langle L(u_m-iu_n) , L(u_m-iu_n) \rangle = \langle u_m-iu_n , u_m-iu_n \rangle = \|u_m-iu_n\|_U^2. $$ Similarly we have: $$ \langle L u_m,i Lu_n \rangle + \langle i L u_n, Lu_m\rangle = \langle u_m , i u_n \rangle + \langle i u_n , u_m \rangle, $$ and this is $$ -i \langle L u_m, Lu_n \rangle + i\langle L u_n, Lu_m\rangle = -i\langle u_m , u_n \rangle + i\langle u_n , u_m \rangle, $$ cancelling $-i$ we have: $$ \langle L u_m, Lu_n \rangle -\langle L u_n, Lu_m\rangle = \langle u_m , u_n \rangle -\langle u_n , u_m \rangle.\tag{4} $$ (3)+(4) gives the result you want.
Following is the if direction proof :(deleted)EDIT 2: Thanks Jonas Meyer pointed out here, the if direction presumes the space has a Schauder basis, i.e., at least separable. By OP only said it is an inner product space, we can't do that.