Looking into textbooks, I got the impression in renormalization perturbation theory one adds counterterms to the Lagrangian to cancel terms (usually integrals) that are infinite.
My question is, could one just ignore the infinite integrals (setting them to zero, so to say), and get the same results as by adding counterterms to cancel them, or does one really get DIFFERENT FINITE RESULTS from the remaining finite terms by adding counterterms?
Renormalization is more than just ignoring singularities, but rather studying how these signularities arise, how they behave, and finding relations between the properties of different particles/objects/systems despite the presence of singularities and independently of them.
Simply removing a singularity introduces a cutoff. We do not call it renormalization, but rather cutting off the singularity or introducing a cutoff. E.g., we can start with an integral $$ I_\alpha = \int_0^{+\infty}\frac{\text{d}x}{x^\alpha} $$ Say $\alpha>1$ - in this case the integral diverges near $0$ and we can cut it off $$ I_\alpha(\Delta) = \int_\Delta^{+\infty}\frac{\text{d}x}{x^\alpha}=\frac{1}{1-\alpha}\frac{1}{x^{\alpha-1}}|_\Delta^{+\infty}= \frac{1}{\alpha-1}\frac{1}{\Delta^{\alpha-1}} $$ The result is obviously dependent on the cutoff $\Delta$, and we would get different numbers depending on the cutoff we choose.
Renormalization would mean studying how this integral diverges (e.g., it diverges at $x\rightarrow 0$ for $\alpha>1$, but at $x\rightarrow+\infty$ for $\alpha>1$, and logarithmically in both limits for $\alpha=1$.) It would also mean studying how it compares to other integrals, and how the denominator influences the value of integral, e.g., $$ \int_0^{+\infty}\frac{f(x)}{x^{\alpha}}\text{d}x $$ Dealing with more complex expressions we could study which terms diverge faster, and therefore more important, e.g., $$ I_\alpha = \int_0^{+\infty}\text{d}x\left(\frac{A}{x^\alpha}+ \frac{B}{x^\beta}\right). $$
This is not unlike studying limits in basic mathematical analysis: $\log(x)$ and $x$ both tend to infinity as $x\rightarrow+\infty$, but what about the ratios $\log(x)/x$ or $x/\log(x)$?
Finally, renormalization is not only about dealing with truly divergent expressions. As Shankar point out in their well-known review, renormalization might have never been invented, if there were no divergences in the QFT... but it turned out to be useful even in the fields there divergencies do not exist, because of the presence of natural cutoffs, such as, e.g., the lattice constant in condensed matter systems. I recommend reading the introduction to this review for getting a brief idea of what renormalization is about, without going too deep into it.
Remarks:
In the minimal subtraction (${\rm MS}$) scheme and the modified minimal subtraction ($\overline{\rm MS}$) scheme, the counterterms only absorb the divergent parts, cf. OP's question. However, there still remains a non-trivial task to calculate the beta functions and analyze the renormalization group flow, i.e. the running of the coupling constants, mass parameters, etc.
