How to prove that in the natural numbers "if $a = b$ then $a + c = b + d$ if and only if $c = d$" using Peano Axioms? I want to prove using the Peano axioms that in the natural numbers if $a = b$ then $a + c = b + d$ if and only if $c = d$ preferably by induction.

Once you assume $a = b$, then showing $c = d \rightarrow a + c = b + d$ is just a matter of using the $=$-rules.
The more interesting part is showing $a + c = b + d \rightarrow c = d$. Since $a = b$, that reduces to showing that $b + c = b + d \rightarrow c = d$, and to show that, you need to use Induction over $b$.
For the base case, this requires proving that $0 + a = a$ for any $a$, which you can do by induction itself. But assuming you have that: $0 + c = 0 + d \rightarrow c = d$ Check
Step: For this, you'll need that (AdditionLeftRecursion) $s(a) + b = s(a + b)$ for any $a$ and $b$, which again can be proven fairly easily by induction itself. But once you have that:
Assume (inductive hypothesis) $b + c = b + d \rightarrow c =d$. Then:
$s(b) + c = s(b) + d \Rightarrow$ (AdditionLeftRecusion) 
$s(b + c) = s(b + d)\Rightarrow$ (Peano 4) 
$b + s(c) = b + s(d) \Rightarrow$ (Inductive Hypothesis) 
$s(c) = s(d) \Rightarrow$ (Peano 2) 
$c = d$
