Entire function bounded at every line

Yes, there are such functions. Take a very narrow region $D$ containing the positive ray, with nice boundary and such that $D$ intersects every any horizontal line other than the real line by a bounded interval. Let $g$ be a conformal map of $D'$ onto the right half-plane, where $D'\subset D$ is another similar region. Then with appropriate choice of $D$ and $D'$ $$f(z)=\int_{\partial D'} \frac{e^{g(\zeta)}}{\zeta-z}d\zeta$$

will converge for $z$ outside $D$ and the function $f$ will be bounded outside $D$ and extend to an entire function (by a deformation of the contour).

For the details, see for example MR2753600 or MR0545054.

EDIT. The method is very flexible of course. Taking $D$ to be a half-strip and $g(z)=e^z$ one obtains a Mittag-Leffler functiuon. Replacing it by $f(z+4i)$ you obtain a function that is bounded on every line from the origin. But to obtain a function as you ask, a half-strip $D$ will not work, so the function is less elementary.

With the same method one can also construct functions which tend to zero on every line: just replace $f$ by $f(z)/(z-z_0)$ where $f(z_0)=0$. Existence of infinitely many zeros of the original $f$ is easy to prove. Repeating this you can find a function which tends to $0$ on every line faster than any polynomial.

Yes. It is even better ! See the friendly paper MR2290290 David H. Armitage, Entire functions that tend to zero on every line. Amer. Math. Monthly 114 (2007), no. 3, 251–256.

Notice that such an example shows that there exist non-trivial functions (any directional derivative of the entire function) in the plane, whose Radon transform vanishes identically. Whence the necessity of rather strong assumptions when inverting the Radon transform. If you think that this inversion is instrumental in the medical scanner, you see that accuracy of mathematical statements can be crucial for humain health!

This should be a comment but it is too long... It is also possible to construct a function satisfying these demands using Taylor series only. For instance, the function

$$F(z)=\sum_{n\geq 0} \frac{z^n}{(\log(\log(n+1+e^e)))^n}$$ is unbounded on the positive real ray, but it is bounded outside the domain $\{x+iy\;:\; x>0,\; |y|<\pi e^{-x}\}$.

To see it, just write $$F(z)=\int_{-1/2-i\infty}^{-1/2+i\infty}\frac{z^s}{(\log\log(s+1+e^e))^s}\frac{ds}{e^{2\pi i s}-1}$$ and shift the contour.