Is every integer a unary operation?

No, it is incorrect to say "an integer is a unary operation". What the other answers so far are actually saying instead is that any integer is the result of repeatedly applying a unary operation, which is quite different.

The true statement I think you are after is that any given integer is (or can be identified with) a nullary operation, specifically the operation taking no arguments and returning itself. See this math.SE thread for example.

The Peano axioms start with a unary operation called $S$ and intended as successor. In fact we define $1=S(0), 2=S(S(0)),$ etc. as $1,2,$ etc. are not part of the language. Then there are axioms that define addition and multiplication to make them work the way they should.

No, an integer is not a unary operation, it is a $0$-ary operation. A unary operation is essentially a function, it requires one input, for example, negation $x\to-x$ is a unary operation. The operation "5" requires zero operands, it has a constant value.

On the other hand, as pointed out in the other answers, the successor operation $x\to x+1$ is unary, and addition is repeated succession.

You've discovered Church numerals. Almost.

In Church encoding, numbers are in fact defined as operations. Actually, a bit more involved: the $n$-th Church numeral is a function that takes a function, and returns the $n$-fold composition of that function with itself. Written in lambda calculus, the first few numbers look thus: $$ 0 = \backslash f\ x\mapsto x $$ $$ 1 = \backslash f\ x\mapsto f\,x $$ $$ 2 = \backslash f\ x\mapsto f(f\,x) $$ $$ 3 = \backslash f\ x\mapsto f\bigl(f(f\,x)\bigr) $$ etc. In other branches of maths, these would be written rather something like $$ 0(f,x) = x $$ $$ 1(f,x) = f(x) $$ $$ 2(f,x) = f\bigl(f(x)) $$ $$ \ldots $$ Now you might ask, are these the natural numbers, or is it just some particular encoding for the naturals? But that's a philosophical question. The church numerals are a model satisfying the Peano axioms, so basically whenever you see numbers used it's possible to put in these definitions, it won't change the meaning.

Yes, you may see this in some more advanced course. Say $S$ is the unary operation "add 1". Then addition may be defined recursively in terms of $S$. $$ x+0 := x; \\ x+Sy := S(x+y) $$

Why not define multiplication next? $$ x*0 := 0 \\ x*Sy := (x*y)+x $$

Can you do powers $x^y$?

Look for "Peano's axioms" for more.