Let $\mathsf{PA_2}$ be a standard recursively axiomatized second-order Peano Arithmetic (e.g. the theory which Hilbert and Bernays call $\mathsf{Z}_2$). And interpret the wffs of $\mathsf{PA_2}$'s language with the 'full' semantics, where the second-order quantifiers are constrained to run over the full collection of sets of numbers.
Then (i) Gödel's theorem applies, of course, to $\mathsf{PA_2}$, so (assuming consistency) there will be a true canonical Gödel sentence $\mathsf{G}$ such that $\mathsf{PA_2} \nvdash \mathsf{G}$.
However, (ii) $\mathsf{PA_2}$, with the full semantics, is categorical by Dedekind's argument, so has only one model (up to isomorphism). In that one model, $\mathsf{G}$ is true, so $\mathsf{PA_2} \vDash \mathsf{G}$.
This shows that syntactic provability $\vdash$ falls short of semantic entailment $\vDash$ in the scecond-order case (there is no complete deductive system for second-order consequence, assuming the full semantics).
In summary, then, while the claim "$\mathsf{PA}$ settles all arithmetical truths" is false however we interpret it (where $\mathsf{PA}$ is first-order Peano Arithmetic), the situation with the corresponding claim "$\mathsf{PA_{2}}$ settles all arithmetical truths" is more complex. It depends whether you mean "semantically entails" or "deductively implies".
For vividness, it may well help to put that symbolically. We'll use $\{\mathsf{PA}, \vdash\}$ to denote the set of theorems that follow in $\mathsf{PA}$'s formal proof system, and $\{\mathsf{PA}, \vDash\}$ to mean the set of sentences semantically entailed by $\mathsf{PA}$'s axioms (given the standard semantics for first-order arithmetic). Similarly, we'll use $\{\mathsf{PA_{2}}, \vdash_2\}$ to mean the set of theorems that follow in $\mathsf{PA_{2}}$'s formal proof system, and $\{\mathsf{PA_{2}}, \vDash_{2}\}$ to mean the set of sentences semantically entailed by $\mathsf{PA_{2}}$'s axioms (given the "full" semantics for the quantifiers). Finally we'll use $\mathcal{T}_{A}$ to denote the set of truths of first order arithmetic (often called, simply, True Arithmetic); and we'll now use $\mathcal{T}_{2A}$, the set of truths of the language of second-order arithmetic (True Second-Order Arithmetic). Then we can very perspicuously display the relations between these sets as follows, using `$\subset$' to indicate strict containment:
$\{\mathsf{PA}, \vdash\} \;=\; \{\mathsf{PA}, \vDash\} \;\subset\; \mathcal{T}_{A}$
$\{\mathsf{PA_{2}}, \vdash_2\} \;\subset\; \{\mathsf{PA_{2}}, \vDash_{2}\} \;=\; \mathcal{T}_{2A}$.
The completeness of first-order logic which yields $\{\mathsf{PA}, \vdash\} = \{\mathsf{PA}, \vDash\}$ is, though so familiar, a remarkable mathematical fact which is enormously useful in understanding $\mathsf{PA}$ and its models. This has led some people to think the strict containment $\{\mathsf{PA_{2}}, \vdash_2\} \subset \{\mathsf{PA_{2}}, \vDash_{2}\}$ is a sufficient reason to be unhappy with second-order logic. Others think that the categoricity of theories like second-order arithmetic, and the second-order definability of key mathematical notions like finiteness, suffices to make second-order logic the natural logic for mathematics. Stewart Shapiro's Foundations without Foundationalism is a classic defence of the second position.
The issue here is discussed at some length in Ch. 29 (or Ch. 22 of the 1st edition) of my Gödel book, if you want more!
Write the axioms of number theory (called "Peano arithmetic," or "PA") as $P^-+\mathrm{Ind}$, where $P^-$ is the ordered semiring axioms (no induction), and $Ind$ is the axiom (scheme) of induction. Then a theorem requires some induction if it is not provable by $P^-$ alone - that is, if we can find a model of $P^-$ in which the theorem is not true.
So that lets us make a precise "Question 1:"
Question 1: Is there a "natural" theorem about natural numbers which requires some induction, in the sense above?
We can refine this. Let's suppose we want to draw a line between "simple" proofs by induction, and "hard" proofs by induction; we allow the former but are skeptical about the latter.
In this case, in the same way that we broke the axioms of number theory up into parts ($P^-$ and $Ind$), we now need to break $Ind$ up into smaller pieces. The usual way of doing this is via the arithmetical hierarchy: every formula in the language of arithmetic can be assigned a "level of complexity," and these levels are indexed by natural numbers. $Ind$ can now be written as $\mathrm{Ind}_1+\mathrm{Ind}_2+ \cdots$, where $Ind_n$ is induction for formulas of complexity $n$. Roughly speaking, a formula has complexity $n$ if it can be written with $n$ alternating blocks of quantifiers: e.g., the formula $$p(a)=\text{“ }\forall x\,\forall y\,\exists z\,\forall w\,(x+y+w<z+a)\text{ ''}$$ has complexity 3, since it has the form "$\forall\forall, \exists, \forall$." (I'm being very vague here, and this is slightly incorrect; but it won't cause any problems.)
So we have question 2:
Question 2: For each $n$, are there "natural" theorems about natural numbers which require $\mathrm{Ind}_n$?
Note that if a theorem requires $\mathrm{Ind}_n$, then it certainly requires some induction; so question 2 is a strengthening of question 1. By the way, this is of course not the only way to break up $\mathrm{Ind}$; there are lots of other ways to measure the complexity of an axiom of induction.
The answers to both questions are, spectacularly, YES; the general question, "How much induction do we need to prove $\varphi$?" is studied - along with similar questions - by the field Reverse Mathematics.
Some examples:
The statement "There are infinitely many primes" is not provable in $P^-$ alone; that is, it requires some induction. How much induction exactly? I don't think this is known, but the (very weak) level of induction called open induction is known to also not be enough.
Ramsey's theorem for pairs - the statement, "Any time I color pairs of natural numbers 'Red' or 'Blue,' I can find an infinite set of natural numbers, any pair from which is colored the same as any other pair" - requires some induction. Again, exactly how much is not known, but it's at least $\mathrm{Ind}_2$ - a small but substantial amount of induction. EDIT: I'm being somewhat sloppy here. Note that Ramsey's theorem isn't expressible in the language of number theory, so I need to look at a more expressive theory which can talk about such things; the theory used for this purpose is usually $RCA_0$, which corresponds in a particularly nice way to the first level of induction + the ability to talk about sets. See Simpson's book (mentioned below) for details on this.
A lot of algebraic statements, like "Every ring which is not a field, has a nontrivial proper ideal", actually require all the induction that $PA$ has to offer.
REFERENCES
Since there's a lot of stuff here, I don't have time to give a complete explanation - but here are some sources:
For basic logic, including Gödel's completeness theorem and what a "model" is, I'm a fan of Enderton's "A Mathematical Introduction to Logic" - but there are lots of books out there on the same subject, and any of them will do.
For the arithmetical hierarchy, this will be covered in any good logic textbook, but can also be found in a lot of books on theoretical computer science - I'm pretty sure it's in Arora/Barak, for example.
For reverse mathematics, this is trickier; there isn't really any readable introduction. The classic text is Simpson's "Subsystems of Second-Order Arithmetic," and chapter I is very nice and readable, but the rest is very hard.
Best Answer
You initially ask:
The answer is basically, yes.
There is very little that can be proved genuinely without induction. Robinson arithmetic, for example, cannot prove that addition is commutative, or that every number is either even or odd. More subtly, it cannot even prove that "$\sum_{i=1}^ni$" is always well-defined - that is, it cannot prove the statement "for all $n$ there is a unique sequence $s$ of length $n$ such that $s_1=1$ and for each $1<k\le n$ we have $s_k=s_{k=1}+k$" (the final term $s_n$ would then be $\sum^n_{i=1}i$). Neither can Robinson arithmetic prove that every bounded subset of $\mathbb{N}$ is finite (which again has to be phrased in terms of sequences to even be expressible in the language of arithmetic).
The main "freedom" which dropping induction entirely permits is the existence of computable models (see Tennenbaum's theorem). For example, the set of polynomials in one variable with integer coefficients and positive leading term (fine, also including $0$) forms a computable nonstandard model of Robinson arithmetic which is fairly interesting in its own right. However, nobody would propose it as a satisfactory universe within which mathematics can properly be conducted (since for example it can't make sense of basic combinatorics).
That said, dropping most induction is more interesting. The system $I\Sigma_1$ (= the ordered semiring axioms + induction for $\Sigma_1$ formulas) is drastically weaker than PA but still proves many of the results of classical number theory. Additionally, it's strong enough that we can meaningfully ask how much additional induction is needed to prove the things it can't - this is part of reverse mathematics (which also studies the role of set existence axioms and their computability-theoretic content in proving higher-type statements), and I've said a bit about this here. Robinson arithmetic is just too weak for "reverse mathematics over Robinson arithmetic" to work properly (although we can go substantially below $I\Sigma_1$).
Meanwhile, the situation in set theory is dramatically different. ZF (= set theory without choice) can still prove a huge amount of mathematics, so it's "usable" in a way that Robinson arithmetic is not. Additionally, there are natural principles contradicting choice, so there's a positive reason to consider ZF; by contrast I don't know of any natural arithmetic principles contradicting induction (although it would be really cool to find some).