$f$ is convex iff $f(x)-xf'(x)$ is decreasing, if $x>0$ $f$ is convex iff $f(x)-xf'(x)$ is decreasing, if $x>0$.
How to prove this, if we just assume $f$ is just differentiable!
The $\Rightarrow$ part is easy. How to prove the $\Leftarrow$ part?
Let $F(x)=f(x)/x$, then $x^2F'(x)$ is increasing. and for $x>0$, $F'(x)>0, F$ increasing. Is this help?
The $\Rightarrow$ part. Since $f$ is convex, we have $f'\nearrow$. For $x<y$,
$$f(x)-xf'(x)=f(y)+f'(\xi)(x-y)-xf'(x)
\geq f(y)+f'(y)(x-y)-xf'(x)
=f(y)-yf'(y)+x[f'(y)-f'(x)]\geq f(y)-yf'(y).$$
Hence, $f(x)-xf'(x)\searrow$.

A differentiable function is convex if and only if $f'$ is increasing, therefore we can formulate the problem as follows:

Let $f:(0, \infty) \to \Bbb R$ be differentiable. Then $f'$ is increasing if and only if $x \mapsto xf'(x) -f (x)$ is increasing.

This can indeed be proven without additional assumptions on the derivative. I'll denote $h(x) = xf'(x) -f (x)$.
Proof of $\implies$: This is similar to what you did. Assume that $f'$ is increasing. For $0 < x < y$ is
$$
 h(y)-h(x) = yf'(y)-xf'(x) - f(y)+f(x) \, .
$$
Using the mean-value theorem we get that for some $c \in (x, y)$
$$
h(y)-h(x) = yf'(y)-xf'(x) - (y-x)f'(c) \\ = x(f'(c)-f'(x)) + y(f'(y) - f'(c)) \ge 0 \, .
$$
This shows that $h$ is increasing.
Proof of $\impliedby$: Assume that $h$ is increasing. Using
$$
 \left( t f(\frac 1t)\right)' = f(\frac 1t) - \frac 1t f'(\frac 1t) = -h(\frac 1t) 
$$
and the mean-value theorem we get for $0 < x < y$, with some $c \in (x, y)$:
$$
 f'(\frac 1y) - f'(\frac 1x) = y h(\frac 1y) - x h(\frac 1x) + y f(\frac 1y) - x f(\frac 1x) \\
= y h(\frac 1y) - x h(\frac 1x) + (y-x) (-h(\frac 1c)) \\
= x \left( h(\frac 1c)-h(\frac 1x)\right) + y \left( h(\frac 1y)-h(\frac 1c)\right) \le 0 \, .
$$
This shows that $x \mapsto f'(1/x)$ is decreasing, so that $f'$ is increasing.
Remark: The proof also shows that

$f'$ is strictly increasing if and only if $x \mapsto xf'(x) -f (x)$ is strictly  increasing.

