Intuitively, "$f(x)$ tends to infinity with $x$" means that as $x$ gets large, $f(x)$ gets larger than any number you pick. Because $f(x)$ is monotonic, once it gets larger than something, it stays larger than that something.
The definition is how to prove it. Say you claim that $f(x)$ goes to infinity. I am allowed to challenge you with some number $B$. To meet the challenge you have to respond with some number $A$ that guarantees if $x \gt A,$ then $f(x) \gt B$. This is a translation of the $\epsilon - \delta$ proof of a limit to "numbers near $\infty$.
For two examples, let $f(x)=\frac x{1000}$ If I challenge you with $B=1,000,000$, you can answer $A=1,000,000,000$ and demonstrate that if $x \gt 10^9, f(x) \gt 10^6$. You can even show that whatever $B$ I name, you win if $A=1000B$. Let $g(x)=10^9-\frac 1x$. If I challenge you with $B=10^6$, you can win with any $A \gt 0$. But if I am clever enough to challenge you with $B=10^{12}$ you lose. So $g(x)$ doesn't go to $\infty$ with $x$.
You might look at Jared's answer to the problem you were given.
Added: Your questioner has told you that $f(x)$ goes to infinity with $x$, given you a $y \gt f(a)$ and asked you to find an $x$ such that $f(x)=y$ and prove it unique. To prove it exists, challenge him with $B=2y$ and he will give you an $A$ such that if $z \gt A, f(z) \gt 2y$. Now the intermediate value theorem shows there is such an $x$ in $[a,A]$ and monotonicity shows it is unique.
Recall that if $f$ is invertible, then it is bijective. So, there exists a unique $y$ such that $y = f(x)$ on the domain of $g$, which is seen to be $(f(a),f(b))$ since we have an increasing function. By definition of the derivative:
$$g'(y) = \lim_{z \rightarrow f(x)} \frac{g(z) - g(f(x))}{z-f(x)} = \lim_{z \rightarrow f(x)} \frac{g(z) - x}{z-f(x)}$$
Now, as we tend $z$ closer and closer to $f(x)$, eventually it will have to belong to $(f(a),f(b))$, which means we can find another $x_z$ such that $f(x_z) = z$. We now choose $z$ sufficiently close, and take advantage of this fact. We then have:
$$g'(y) = \lim_{z \rightarrow f(x)} \frac{g(f(x_z)) - x}{f(x_z)-f(x)} = \lim_{z \rightarrow f(x)} \frac{x_z - x}{f(x_z)-f(x)}$$
We see that this final limit tends to $\frac{1}{f'(x)}$.
Best Answer
A necessary and sufficient condition is that $f^\prime\ge0$ everywhere and $f^\prime > 0$ on a dense subset of $\mathbb{R}$. The condition stated in the question is sufficient, but not necessary. Consider the following differentiable and strictly increasing function for which $f^\prime > 0$ does not hold almost everywhere. Let $C$ be the fat Cantor set (which is closed with positive measure, but does not contain any nontrivial open intervals). Let $g(x)$ be the distance from any point $x$ to $C$. Then, $g$ is continuous and vanishes precisely on $C$. The integral $$ f(x)=\int_0^xg(y)\,dy $$ is strictly increasing and continuously differentiable, but $f^\prime=0$ on $C$.
We can show that $f^\prime\ge0$ everywhere and $f^\prime > 0$ on a dense set is indeed necessary and sufficient for $f$ to be strictly increasing (see also Olivier Bégassat's comments to the question).
Sufficiency: If $f^\prime\ge0$ then the mean value theorem implies that $f$ is increasing. If it was not strictly increasing then we would have $f(a)=f(b)$ for some $a < b$ implying that $f^\prime=0$ on $[a,b]$, contradicting the condition that $f^\prime > 0$ on a dense set. This proves sufficiency of the conditions.
Necessity: In the other direction, suppose that $f$ is strictly increasing. Then, $f^\prime\ge0$ follows directly from the definition of the derivative. Also, $f(a) < f(b)$ for any $a < b$. The mean value theorem implies that $f^\prime(x) > 0$ for some $x\in[a,b]$. So, $f^\prime > 0$ on a dense set.