$f$ 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: 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.

A: Proof of Equivalence
Suppose $f(x)-xf'(x)$ is decreasing. For $\Delta x\gt0$,
$$
f(x)-xf'(x)\ge f(x+\Delta x)-(x+\Delta x)f'(x+\Delta x)\tag1
$$
For $\Delta x\lt0$, inequality $(1)$ is reversed. In either case, dividing by $\Delta x$ and rearranging, we get
$$
(x+\Delta x)\,\frac{f'(x+\Delta x)-f'(x)}{\Delta x}\ge\frac{f(x+\Delta x)-f(x)}{\Delta x}-f'(x)\tag2
$$
Dividing by $x+\Delta x$ and taking the $\liminf$, we get
$$
\liminf_{\Delta x\to0}\frac{f'(x+\Delta x)-f'(x)}{\Delta x}\ge0\tag3
$$
Thus, as shown in this answer, $f'(x)$ is increasing.

Suppose $f'(x)$ is increasing. For $\Delta x\gt0$,
$$
x\,\frac{f'(x+\Delta x)-f'(x)}{\Delta x}\ge0\ge\overbrace{\frac{f(x+\Delta x)-f(x)}{\Delta x}}^{f'(x+t\Delta x),\,t\in(0,1)}-f'(x+\Delta x)\tag4
$$
Rearranging and multiplying by $\Delta x$, we get
$$
f(x)-xf'(x)\ge f(x+\Delta x)-(x+\Delta x)f'(x+\Delta x)\tag5
$$
Thus, $f(x)-xf'(x)$ is decreasing.

Geometric Representation
$f(x)-x\,f'(x)$ is the $y$-intercept of the tangent to the graph of $f$:

