Is $Ax\cdot y=x\cdot A^\dagger y$ true for any scalar product? Let $A\in M(n\times n, \mathbb{C})$ and $V$ be a unitary space with $dim(V)=n$. How can I prove for any $x,y\in V$ $$Ax\cdot y=x\cdot A^\dagger y?$$
I managed to show that it works with $V=\mathbb{C^n}$ and $x\cdot y=x_j y^*_j$. However I had to use my definition of scalar product. Can you give me a hint on how to prove this for any scalar product, if it is possible?
 A: "Yes and no".
For any linear operator $A$ on your abstract unitary space $V$ with given scalar product $(. , .)$ you can define the linear operator $A^*$ as the unique operator satisfying $(A^*v, w) = (v, Aw)$ for all $v, w \in V$. (You should think a bit about why this definition actually makes sense and why $(A^*)^* = A$.)
Now given a basis $\mathcal{B}$ of $V$ we can write the matrix (as in 'block of numbers') $A_\mathcal{B}$ representing operator $A$. This thing comes with a matrix  $(A_\mathcal{B})^\dagger$, which I guess you define as the conjugate transpose.
Now your question can be reworded as:
Is the matrix $(A_\mathcal{B})^\dagger$ the same as the matrix $(A^*)_\mathcal{B}$ representing operator $A^*$ in the same basis?
The answer is: 'if and only if $\mathcal{B}$ is an orthonormal basis'.
(Note that the notion orthonormal basis makes sense because you have defined a scalar product on $V$, but also that whether or not a given basis is orthonormal depends on the choice of scalar product.)
EDIT: This stuff is easier to understand when you have an example of an inner-product space which has a way to give every vector a unique name that is not a list of $n$ numbers. One example is the space $V$ of all functions $f: [0, 1] \to \mathbb{C}$ of the form $f(x) = ax + b$ with inner product $(f, g) = \int_0^1 f(x)\overline{g(x)} dx$.
Let functions $f_1, f_2, f_3 \in V$ be given by $f_1(x) = 1$, $f_2(x) = x$ and $f_3(x) = \sqrt{3} (2x - 1)$. 
The reason for introducing these is that the naive basis $\mathcal{N} = \{f_1, f_2\}$ is not an o.n.b. and the fancy basis $\mathcal{B} = \{f_1, f_3\}$ is.
Now we need an example of a linear operator on this space. One is the operator $A$ defined by $(Af)(x) = f(1-x)$. You can check that it is unitary w.r.t. the inner product.
Now the matrix $A_\mathcal{N} = \begin{pmatrix} 1 &  1 \\ 0 & - 1 \end{pmatrix}$ while the matrix $A_\mathcal{B} = \begin{pmatrix} 1 &  0 \\ 0 & - 1 \end{pmatrix}$.
In this very special case we have that $A^* = A$ which, concretely means that for all $f, g \in V$ we have that $\int_0^1 f(x)\overline{g(1-x)}dx = \int_0^1 f(1-x)\overline{g(x)}dx$. we see that in matrix-land this corresponds to $A_\mathcal{B}^\dagger = A_\mathcal{B}$ but not so much to $A_\mathcal{N}^\dagger = A_\mathcal{N}$.  
FURTHER UPDATE: after staring at this example I get the impression that being an orthogonal basis (that is $(v, w) = 0$ for $v, w$ different basis elements) is enough, and we do not need the full force of the basis being orthonormal (that is: orthogonal and $(v, v) = 1$ for each basis element.) Please check this yourself and let me know!
YES! IT IS TIME FOR ANOTHER UPDATE! 
So, what the above example shows is why, when translating linear operators into matrices by picking a basis, it is necessary to pick an orthogonal basis. What is not so clear (see comment by OP) is why it is also sufficient.
So here is the reason. Picking a basis not only assigns a matrix (block of numbers) to every linear operator on your vector space $V$ but also a column vector in the sense of list of numbers to every vector in $V$. Write $v_\mathcal{B}$ for the column vector representing the 'abstract' vector $v$. (In the example: $(f_2)_\mathcal{B} = \begin{pmatrix} 1/2 \\ \sqrt{3}/6 \end{pmatrix}$.)
Now suppose that we have chosen an orthonormal basis $\mathcal{B}$ ov $V$ (I stick with this case because it is easier, you should think a bit about how to modify the argument in the orthogonal case), where orthonormal is defined w. r. t. the inner product $(. , .)$ on $V$. For every column vector (as in list of numbers) $a$ we write $a^\dagger$ for its conjugate transpose, which is a row vector.
The thing you need to check is the following: for every two vectors $v, w \in V$ we have that $$(v, w) = (v_\mathcal{B})^\dagger w_\mathcal{B}.$$
Here on the left hand side we have a number, because that is what the inner product $(., .)$ produces, and on the right hand side we have a number because we multiply a row vector with a column vector yielding a 1-by-1-matrix that can and should be re-interpreted as a number.
Once you have verified this statement you are essentially done because checking the original statement $(A_\mathcal{B})^\dagger = (A^*)_\mathcal{B}$ then reduces to checking the case that you say in the original post that you have already checked!
FINAL REMARK ON NOTATION: I use $\dagger$ when operating on matrices and $*$ when operating on linear operators because for the purposes of this answer it is useful to keep the distinction between the two worlds as clear as possible. However in practice both symbols are used in both situations.
