$a < b$ and $c<d$ imply $a+c < b+d$ when $a,b,c,d$ are arbitrary nonnegative integers.
I know that (assuming we include zero) $$\begin{align*} a<b \Leftrightarrow (\exists x\in \mathbb N)a + S(x) = b\\ c<d \Leftrightarrow (\exists y\in \mathbb N)c + S(y) = d \end{align*}$$
And that's what I have done by using associative and conmutative properties:
$$(a+c)+(S(x)+S(y))\\=a+(c+S(x))+S(y)\\=a + (S(x)+c)+S(y)\\=(a+S(x)) + (c+S(y))\\= b + d$$
Also, we have that $S(x) + S(y) = S(S(x+y))$ by using addition definition recursively.
$x \land y \in \mathbb{N} \implies S(x) \land S(y) \in \mathbb{N}^*$ where $\mathbb N^* =\mathbb N\setminus\{0\}$
That means we could start with $(a + c) + 1 = b + d$ and could work with any $k \in \mathbb N^*$ that satisfies the expression for arbitrary $a, c, b, d \in \mathbb N$, thus $a+c < b+d$.
Is that proof valid in the context of naturals and Peano axioms?
As pointed out in the comments, your proof is not quite complete.
In order to tshow that $a + c < b + d$, you need to show that there is something $z$ such that $$(b + d) + S(z) = a + c$$
all you have done is to show that
$$(b + d) + (S(x) + S(y)) = a + c$$
Fortunately, this problem is easily rectified, since you can show that $$S(x) + S(y) = S(S(x) + Y)$$
And hence you have the $z$ as need: $S(x) + y$