2 Pointed out a serious error in my answer.

There is a classic paper by Jordan and von Neumann where they prove results that allows this question is settled in an elementary way.

On Inner Products in Linear, Metric Spaces Author(s): P. Jordan and J. V. Neumann, The Annals of Mathematics, Second Series, Vol. 36, No. 3 (Jul., 1935), pp. 719-723.

They first prove by elementary arguments (their Theorem I) the so-called Jordan v. Neumann criterion, that a Banach space is Hilbert iff for all $x$ and $y$, $(*) ||x + y||^2 - ||x - y||^2 = 2||x||^2 + 2 ||y||^2$. They then show from this that a Banach space is Hilbert iff every 1 and 2 dimensional subspace is Euclidean. Here is their argument:

4.The condition that every $<= 2$-dimensional subspace $L'$ of $L$ be isometric to a Euclidean space, is obviously necessary for the existence of an inner-product in the generalized linear,metric space L. It is sufficient,too, because if it is fulfilled, one can argue as follows:If $f_o,g_0\in L$ the space $L'$ of all $\alpha f_0 + \beta g_0$ ($\alpha,\beta$ arbitrary complex numbers) is $<= 2$ dimensional,thus (*) holds in $L'$ (as in every Euclidean space). Therefore it holds in particular for $f = f_0, g = g_0$,and as $f_0,g_0$ are arbitrary,Theorem arbitrary, Theorem I proves the existence of an inner product.

[SEE BELOW: The following sentence does NOT complete the proof !]
And as rpotrie has pointed out in another answer, the two dimensional case follows from the assumed transitivity condition.

ERROR NOTICE: I noticed a serious error in the above reasoning! If the isometry group $G$ of a Banach space $V$ is transitive on the unit sphere of $V$, it does NOT follow in any obvious way that the isometry group of a subspace $V'$ of $V$ is transitive on the unit sphere of $V'$. (If $e_1,e_2$ are unit vectors in $V'$, then an element $g$ of $G$ that carries $e_1$ to $e_2$ need not leave $V'$ invariant.)

I did not at first realize how remarkable the conclusion is that transitivity on the unit sphere implies Euclidean. It can be rephrased as saying that transitivity on $S$ implies $2$-transitivity, which to me at least seems even more remarkable. (It was realizing this fact that let me see my silly error.)

1

There is a classic paper by Jordan and von Neumann where they prove results that allows this question is settled in an elementary way.

On Inner Products in Linear, Metric Spaces Author(s): P. Jordan and J. V. Neumann, The Annals of Mathematics, Second Series, Vol. 36, No. 3 (Jul., 1935), pp. 719-723.

They first prove by elementary arguments (their Theorem I) the so-called Jordan v. Neumann criterion, that a Banach space is Hilbert iff for all $x$ and $y$, $(*) ||x + y||^2 - ||x - y||^2 = 2||x||^2 + 2 ||y||^2$. They then show from this that a Banach space is Hilbert iff every 1 and 2 dimensional subspace is Euclidean. Here is their argument:

4.The condition that every $<= 2$-dimensional subspace $L'$ of $L$ be isometric to a Euclidean space, is obviously necessary for the existence of an inner-product in the generalized linear,metric space L. It is sufficient,too, because if it is fulfilled, one can argue as follows:If $f_o,g_0\in L$ the space $L'$ of all $\alpha f_0 + \beta g_0$ ($\alpha,\beta$ arbitrary complex numbers) is $<= 2$ dimensional,thus (*) holds in $L'$ (as in every Euclidean space). Therefore it holds in particular for $f = f_0, g = g_0$,and as $f_0,g_0$ are arbitrary,Theorem I proves the existence of an inner product.

And as rpotrie has pointed out in another answer, the two dimensional case follows from the assumed transitivity condition.