Yes, $f(x)=\sin(x)$. The proof below is not entirely elementary. There's probably a more elementary proof, although nobody's come up with one yet; the proof below does tell us a few interesting things about the class of functions satisfying the given condition.

Say $f\in X$ if $f$ is smooth and $$||f||=||f||_X=\sup_{k\ge0}\sup_{x\in\Bbb R}|f^{(k)}(x)|<\infty.$$ (An example of one of the interesting things mentioned above: We will see below that if $f\in X$ then $||f||_X=||f||_\infty$.)

As has been noted, if $f\in X$ then Taylor's Theorem with a suitable form of the remainder shows that $f$ is equal to its Maclaurin series on all of $\Bbb R$; thus $f$ is (the restriction to $\Bbb R$ of) an entire function. Simply noting $|\sum a_nz^n|\le\sum|a_n|\,|z|^n$ shows that $|f(z)|\le||f||e^{|z|}$. In particular $|f(iy)|\le||f||e^{|y|}$; now since $X$ is translation-invariant it follows that $$|f(x+iy)|\le||f||e^{|y|}.$$

Functions in $X$ enjoy a certain magical property:

**Magical Property** If $f\in X$ then $$f'(x)=\sum_{k=-\infty}^\infty
\frac{4(-1)^k}{(2k+1)^2\pi^2}f(x+\pi(k+1/2)).$$

(Hence $||f'||_\infty\le||f||\infty$ as noted above.)

The proof of the Magical Property is the not quite so elementary part of the argument (maybe it follows by some elementary complex analysis, I don't see how). First we show how the result follows:

Suppose as in the OP that $||f||_X\le 1$, $f$ is real-valued on $\Bbb R$, and $f'(0)=1$. Then $$1=f'(0)
=\sum_{k=-\infty}^\infty
\frac{4(-1)^k}{(2k+1)^2\pi^2}f(\pi(k+1/2)).$$Now since $$\sum_{k=-\infty}^\infty
\frac{4}{(2k+1)^2\pi^2}=1$$and $-1\le f\le 1$ we cannot have any cancellation here; we must have $$f(\pi(k+1/2))=(-1)^k=\sin(\pi(k+1/2)).$$

So if we let $$g(z)=\frac{f(z)-\sin(z)}{\cos(z)}$$then $g$ is an entire function. Since $|f(x+iy)|\le||f||e^{|y|}$ it is clear that $g$ is bounded in the complement of the union of small disks centered at $\pi(k+1/2)$. Hence $g$ is bounded and so $g$ is constant. So $$f(z)=\sin(z)+c\cos(z).$$Since $f$ is real-valued on $\Bbb R$ it follows that $c\in\Bbb R$, and now it's easy to show that $|f|\le1$ implies $c=0$.

**Proof of the Magical Property**

Here we assume the reader is familiar with the basic properties of tempered distributions and their Fourier transforms. If you don't know that stuff probably you shouldn't bother asking about the stuff in this section, it would take a lot of space to explain. Of course if you do know that stuff and something below seems problematic please say so.

Suppose $f\in X$.

Since $f\in L^\infty(\Bbb R)$, $f$ is a tempered distribution, and as such it has a Fourier transform, that being another tempered distribution. A version of the Paley-Wiener Theorem for tempered distributions, in for example Rudin *Functional Analysis*, shows that $$supp(\hat f)\subset[-1,1].$$(This special case is easier than the theorem in Rudin; I may add a few words about that if it looks like anyone's reading this.)

The Magical Property is really just an explicit version of Bernstein's inequality. Recall: If $g\in L^p(\Bbb R)$ and $supp(\hat g)\subset[-1,1]$ then $||g'||_p\le||g||_p$. The proof proceeds by showing that there exists a complex measure $\mu$ with $||\mu||=1$ and $\hat\mu(\xi)=i\xi$ for $-1\le\xi\le 1$; then considering the Fourier transform shows that $g'=g*\mu$, hence $||g'||_p\le||g||_p||\mu||=||g||_p$. Modulo technicalities the Magic Property for $f\in X$ follows by the same argument, except we need to say exactly what $\mu$ *is*.

It's straightforward, if tedious, to verify that $$\sum_{k=-\infty}^\infty\frac{4(-1)^k}{(2k+1)^2\pi^2}e^{i\pi(k+1/2)t}=
\begin{cases}it,&(-1\le t\le 1),
\\i(2-t),&(1\le t\le 3).\end{cases}$$(Simply extend the function on the right to a function of period $4$ and calculate the Fourier coefficients.)
Hence if $$\mu=\sum_{k=-\infty}^\infty\frac{4(-1)^k}{(2k+1)^2\pi^2}\delta_{-\pi(k+1/2)}$$then $$\hat\mu(\xi)=i\xi\quad(-1\le\xi\le 1).$$Since $supp(\hat f)\subset[-1,1]$ this shows that $f'=f*\mu$, which is exactly what the Magic Property asserts.

**A Technicality** Asserting that $f'=f*\mu$ because $\hat\mu(\xi)=i\xi$ on the support of $\hat f$ is a little glib. The problem is that $\hat f$ is just a distribution, and $\hat\mu$ is not smooth on a neighborhood of $[-1,1]$. (This really is a problem; for example if $f$ is an arbitrary tempered distibution with $\hat f$ supported in $[-1,1]$ the convolution $f*\mu$ need not even exist.)

This is not hard to fix. For $0<\lambda<1$ let $$f_\lambda(x)=f(\lambda x).$$ It's enough to show that $f_\lambda'=f_\lambda*\mu$. This is straightforward, since $\hat\mu(\xi)=i\xi$ on a neighborhood of the support of $\hat f_\lambda$. Again, I may add details if it seems worthwhile.

**Amuusing Note** We've actually shown this:

**Theorem** Suppose $f\in L^\infty(\Bbb R)$. TFAE:

$f\in X$.

$f$ extends to an entire function with $|f(x+iy)|\le ce^{|y|}$.

$\hat f$ is supported in $[-1,1]$.

(We've shown explicitly that (1) implies (2) and (2) implies (3). To show (3) implies (1), note that (3) implies that $||f'||_\infty\le||f||_\infty$.)