Since this question is asked often enough, let me add a detailed solution. I'm not quite following Arturo's outline, though. The main difference is that I'm not re-proving the Cauchy-Schwarz inequality (Step 4 in Arturo's outline) but rather use the fact that multiplication by scalars and addition of vectors as well as the norm are continuous, which is a bit easier to prove.
So, assume that the norm $\|\cdot\|$ satisfies the parallelogram law
$$2 \Vert x \Vert^2 + 2\Vert y \Vert^2 = \Vert x + y \Vert^2 + \Vert x - y \Vert^2$$
for all $x,y \in V$ and put
$$\langle x, y \rangle = \frac{1}{4} \left( \Vert x + y \Vert^2 - \Vert x - y \Vert^2\right).$$ We're dealing with real vector spaces and defer the treatment of the complex case to Step 4 below.
Step 0. $\langle x, y \rangle = \langle y, x\rangle$ and $\Vert x \Vert = \sqrt{\langle x, x\rangle}$.
Obvious.
Step 1. The function $(x,y) \mapsto \langle x,y \rangle$ is continuous with respect to $\Vert \cdot \Vert$.
Continuity with respect to the norm $\Vert \cdot\Vert$ follows from the fact that addition and negation are $\Vert \cdot \Vert$-continuous, that the norm itself is continuous and that sums and compositions of continuous functions are continuous.
Remark. This continuity property of the (putative) scalar product will only be used at the very end of step 3. Until then the solution consists of purely algebraic steps.
Step 2. We have $\langle x + y, z \rangle = \langle x, z \rangle + \langle y, z\rangle$.
By the parallelogram law we have
$$2\Vert x + z \Vert^2 + 2\Vert y \Vert^2 = \Vert x + y + z \Vert^2 + \Vert x - y + z\Vert^2 .$$
This gives
$$\begin{align*}
\Vert x + y + z \Vert^2 & = 2\Vert x + z \Vert^2 + 2\Vert y \Vert^2 - \Vert x - y + z \Vert^2 \\
& = 2\Vert y + z \Vert^2 + 2\Vert x \Vert^2 - \Vert y - x + z \Vert^2
\end{align*}$$
where the second formula follows from the first by exchanging $x$ and $y$. Since $A = B$ and $A = C$ imply $A = \frac{1}{2} (B + C)$ we get
$$\Vert x + y + z \Vert^2 = \Vert x \Vert^2 + \Vert y \Vert^2 + \Vert x + z \Vert^2 + \Vert y + z \Vert^2 - \frac{1}{2}\Vert x - y + z \Vert^2 - \frac{1}{2}\Vert y - x + z \Vert^2.$$
Replacing $z$ by $-z$ in the last equation gives
$$\Vert x + y - z \Vert^2 = \Vert x \Vert^2 + \Vert y \Vert^2 + \Vert x - z \Vert^2 + \Vert y - z \Vert^2 - \frac{1}{2}\Vert x - y - z \Vert^2 - \frac{1}{2}\Vert y - x - z \Vert^2.$$
Applying $\Vert w \Vert = \Vert - w\Vert$ to the two negative terms in the last equation we get
$$\begin{align*}\langle x + y, z \rangle & = \frac{1}{4}\left(\Vert x + y + z \Vert^2 - \Vert x + y - z \Vert^2\right) \\
& = \frac{1}{4}\left(\Vert x + z \Vert^2 - \Vert x - z \Vert^2\right) +
\frac{1}{4}\left(\Vert y + z \Vert^2 - \Vert y - z \Vert^2\right) \\
& = \langle x, z \rangle + \langle y, z \rangle
\end{align*}$$
as desired.
Step 3. $\langle \lambda x, y \rangle = \lambda \langle x, y \rangle$ for all $\lambda \in \mathbb{R}$.
This clearly holds for $\lambda = -1$ and by step 2 and induction we have $\langle \lambda x, y \rangle = \lambda \langle x, y \rangle$ for all $\lambda \in \mathbb{N}$, thus for all $\lambda \in \mathbb{Z}$. If $\lambda = \frac{p}{q}$ with $p,q \in \mathbb{Z}, q \neq 0$ we get with $x' = \dfrac{x}{q}$ that
$$q \langle \lambda x, y \rangle = q\langle p x', y \rangle = p \langle q x', y \rangle = p\langle x,y \rangle,$$
so dividing this by $q$ gives
$$\langle \lambda x , y \rangle = \lambda \langle x, y \rangle \qquad\text{for all } \lambda \in \mathbb{Q}.$$
We have just seen that for fixed $x,y$ the continuous function $\displaystyle t \mapsto \frac{1}{t} \langle t x,y \rangle$ defined on $\mathbb{R} \smallsetminus \{0\}$ is equal to $\langle x,y \rangle$ for all $t \in \mathbb{Q} \smallsetminus \{0\}$, thus equality holds for all $t \in \mathbb{R} \smallsetminus \{0\}$. The case $\lambda = 0$ being trivial, we're done.
Step 4. The complex case.
Define $\displaystyle \langle x, y \rangle =\frac{1}{4} \sum_{k =0}^{3} i^{k} \Vert x +i^k y\Vert^2$, observe that $\langle ix,y \rangle = i \langle x, y \rangle$ and $\langle x, y \rangle = \overline{\langle y, x \rangle}$ and apply the case of real scalars twice (to the real and imaginary parts of $\langle \cdot, \cdot \rangle$).
Addendum. In fact we can weaken requirements of Jordan von Neumann theorem to
$$
2\Vert x\Vert^2+2\Vert y\Vert^2\leq\Vert x+y\Vert^2+\Vert x-y\Vert^2
$$
Indeed after substitution $x\to\frac{1}{2}(x+y)$, $y\to\frac{1}{2}(x-y)$ and simplifications we get
$$
\Vert x+y\Vert^2+\Vert x-y\Vert^2\leq 2\Vert x\Vert^2+2\Vert y\Vert^2
$$
which together with previous inequality gives the equality.
$
\newcommand{\norm}[1]{\left\| #1 \right\|}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
$
It so happens that I wrote up a proof of this pretty recently. Here it is, mostly unaltered, except I changed the variables in question to $u$ and $v$. I am using $(u,v)_\C$ here to denote your $\langle u,v \rangle$; I first solve the real case and then the complex case follows directly.
First consider $(u_1 + u_2, v)_\R = \frac{\norm{u_1 + u_2 + v}^2 - \norm{u_1 + u_2 - v}^2}{4}$.
Applying the parallelogram law to each factor, we write
\begin{align*}
\norm{u_1 + u_2 + v}^2 - \norm{u_1 + u_2 - v}^2
&=
\left( 2\norm{u_1 + v}^2 + 2\norm{u_2}^2 - \norm{(u_1 + v) - u_2}^2 \right) \\
&\quad - \left( 2\norm{u_1}^2 + 2\norm{u_2 - v}^2 - \norm{u_1 + (u_2 - v)} \right) \\
&=
2\norm{u_2}^2 - 2\norm{u_1}^2 + 2\norm{u_1 + v}^2 - 2\norm{u_2 - v}^2
\end{align*}
By symmetry (switching $u_1$ and $u_2$) we have the two equations
\begin{align*}
\norm{u_1 + u_2 + v}^2 - \norm{u_1 + u_2 - v}^2
&=
2\norm{u_2}^2 - 2\norm{u_1}^2 + 2\norm{u_1 + v}^2 - 2\norm{u_2 - v}^2
\\
\norm{u_1 + u_2 + v}^2 - \norm{u_1 + u_2 - v}^2
&=
2\norm{u_1}^2 - 2\norm{u_2}^2 + 2\norm{u_2 + v}^2 - 2\norm{u_1 - v}^2
\end{align*}
Averaging (arithmetically) the two equations we vet
\begin{align*}
\norm{u_1 + u_2 + v}^2 - \norm{u_1 + u_2 - v}^2
&=
\norm{u_1 + v}^2 - \norm{u_1 - v}^2
+
\norm{u_2 + v}^2 - \norm{u_2 - v}^2
\end{align*}
which when divided by four is
$$
(u_1 + u_2,v)_\R = (u_1,v)_\R + (u_2,v)_\R.
$$
This is just the real case. We now need to show it is true in the complex case also.
\begin{align*}
(u_1 + u_2,v)_\C
&= (u_1 + u_2,v)_\R + i(u_1 + u_2, iv)_\R \\
&= (u_1,v)_\R + (u_2,v)_\R + i(u_1,iv)_\R + i(u_2,iv)_\R \\
&= (u_1,v)_\C + (u_2,v)_\C
\end{align*}
Best Answer
The paralelogram law holds if and only if the norm is induced by an inner product (over characteristic $\ne 2$):
Supposing the paralelogramma law,
Let $\langle a,b\rangle:=\displaystyle\frac12\left(\|a+b\|^2 - \|a\|^2-\|b\|^2\right)$.
For $\Bbb K=\Bbb C$, we can define a hermitian inner product:
Let $\langle a,b\rangle:=\displaystyle\frac14\left(\|a+b\|^2+i\|a+ib\|^2-\|a-b\|^2-i\|a-ib\|^2\right)$.