Let $g:[0,\infty) \rightarrow \mathbb{R}$ be an increasing function. Is there a way to construct an entire function $f(z)$ such that $f(x)=g(x)$ for all real $x$?

You should keep in mind analytic continuation. Let $f:\mathbb R\rightarrow\mathbb R$ be a $C^\infty$ function ; if there exists an entire function extending $f$, then (1) It is unique by analytic continuation, (2) The function $f$ must be realanalytic. Of course (2) is not sufficient: think about the realanalytic $x\mapsto\frac{1}{1+x^2}$, which does NOT have an entire extension, since by analytic continuation, that extension should coincide with $\mathbb C\ni z\mapsto \frac{1}{1+z^2}$, which has poles at $\pm i$. For a realanalytic $f$, you can formulate a criterion dealing with the size of derivatives. Such a function has an entire extension iff $$\forall R>0,\exists C_R,\forall n\in\mathbb N,\quad \vert f^{(n)}(0)\vert\le C_R\frac{n!}{R^n}. $$ On the other hand, realanalyticity of a $C^\infty$ $f$ on the real line is equivalent to $$ \forall x\in \mathbb R,\exists r>0,\exists C>0, \exists R>0,\forall n\in\mathbb N,\quad \Vert f^{(n)}\Vert_{L^\infty(B(x,r))}\le C\frac{n!}{R^n}. $$ 


You certainly need some smoothness condition for this to be possible. In fact, the condition is that $f$ be extendable to an even, realanalytic function on the line which kind of makes the result a tautology since then $f$ has the form $\sum_{n=1}^\infty a_{2n} z^{2n}$. 

