$f(x)=x^{3}+1$ - Injective and Surjective?

As in the answer you linked to (my answer), I would try to structure your proofs more clearly:

Claim: The mapping $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^3+1$ is injective.

Proof. Let $x_1,x_2\in\mathbb{R}$ and suppose $f(x_1)=f(x_2)$. Then $$ f(x_1)=f(x_2)\\[0.5em] x_1^3+1=x_2^3+1\\[0.5em] x_1^3=x_2^3\\[0.5em] x_1=x_2. $$ Thus, the mapping is injective. $\blacksquare$

Claim: The mapping $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^3+1$ is surjective.

Proof. Suppose $y\in\mathbb{R}$. Then let $x=\sqrt[3]{y-1}$. We have the following: \begin{align} f(x) &= x^3+1\\[0.5em] &= (\sqrt[3]{y-1})^3+1\\[0.5em] &= (y-1)+1\\[0.5em] &= y. \end{align} Thus, the mapping is surjective. $\blacksquare$

Note: As you have seen, writing up proofs in math is not only about squeezing out an accurate proof. You want to write it clearly, concisely, etc. The goal is to be concise but not at the cost of clarity. Oftentimes this is a fine line to walk.

Your proof of of $f$ being injective is correct (although when you are at the point $a^3 = b^3$ you want to raise each side to the $\frac{1}{3}$ power, not the $-3$rd power.) As for the surjective part, I would also personally mention that you know $x = \sqrt[3]{y-1}$ is in the domain of $f$ because $\sqrt[3]{y-1}$ exists for every $y \in \mathbb{R}$. After all, this is the key point of a function being surjective. Not just the fact that you can plug in $x$ and get $y=y$. It may be clear to you or others already that $\sqrt[3]{y-1}$ exists for every $y \in \mathbb{R}$, but if you are just starting to work with into/onto properties of functions, I think it is something that should be explicitly stated. All in all, nice job though!