What is the smallest set of letters that can spell any integer?

  ...
-3: MINUS ONE MINUS ONE MINUS ONE
-2: MINUS ONE MINUS ONE
-1: MINUS ONE
 0: ONE MINUS ONE
 1: ONE 
 2: ONE MINUS MINUS ONE
 3: ONE MINUS MINUS ONE MINUS MINUS ONE
    ...

two letters less since we do not use PLUS any more.

If we use LESS to indicate subtraction, then we have

-1 = ONE LESS ONE LESS ONE

0 = ONE LESS ONE

1 = ONE

2 = ONE PLUS ONE

etc., using only seven letters ELNOPSU.

If any integer can be spelled with this structure

one some operation one some operation one

the trick could be simply this: reduce the numbers of letters needed for operations (or use O, N, E as much as you can!).

M,I,N,U,S introduces 4 new letters (N does not count) P,L,U,S introduces 2 other letters (P and L)

A simple (but symbolic) solution is to represent a operation with a single letter... but this is not a way to "say it in English".

So I propose this structure: PLUS becomes AND (only two more letters, A and D). Every integer number N bigger than 1 can be said as

N =  ONE AND ONE AND ONE AND ONE...

B) Define the number MINUS ONE as "NOT ONE" (let's say this will became our convention, this is the weakest part!) in order to introduce only one more letter (T)

Now zero becomes

ZERO =  ONE AND (NOT ONE)

C) Since every negative numbers is equal to its absolute value multiplied by -1 e.g. -59 = 59 * (-1) I suggest to use a new word to introduce the operation of multiplication, as in class we often say A multiplied by B as "A dot B" (A dot is one possible symbol to show multiplication in advanced math, instead of x or * or others)

Note that now DOT does not introduce any new letter. So -3 could be

(ONE AND ONE AND ONE) DOT (NOT ONE)

Notice we have used only 6 letters instead of 9 (and mantaining a speakable structure) O, N , E, D, T, A

Now the last problem: how to "say" parenthesis? With the rithm of speech! ONE AND ONE AND ONE AND ONE [after this sequence do a little pause] DOT [say DOT as an important word, then a little pause] NOT ONE [say NOT ONE fast, as if would be a single word]

Thank you for this pretty puzzle and sorry for my poor english!

Another way of using only seven letters (A, D, E, L, N, O, S):

    ...
-3: ONE LESS ONE LESS ONE LESS ONE LESS ONE
-2: ONE LESS ONE LESS ONE LESS ONE
-1: ONE LESS ONE LESS ONE
 0: ONE LESS ONE
 1: ONE
 2: ONE AND ONE
 3: ONE AND ONE AND ONE
 4: ONE AND ONE AND ONE AND ONE
    ...

4 letters: E N O T

Explanation:

My idea is to use binary and marcoresk's idea of using NOT to represent the - sign, although I have to admit it might be slightly 'cheating' as I have to define things and work in a different base.

To represent a number you read out its binary representation using ONE and NONE to represent $1$s and $0$s.

Eg. $5$ would be: ONE NONE ONE

If the number is negative, add a NOT to the beginning.

Eg. $-6$ would be: NOT ONE ONE NONE

Extension:

You can represent any decimal which is terminating in binary with an extra letter D for a total of 5 letters which allows you to say DOT for the decimal point.

Eg. $-5.5$ would be: NOT ONE NONE ONE DOT ONE

Since

"O" is in the Collins English dictionary as "used to mean zero"
(pronounced the same as "owe", as in "I owe MackTuesday a coconut")

...and, as pointed out by @MackTuesday, "a" means one
(as in "I owe MackTuesday a coconut")

We can

Use just two letters: O, and a
(I previously used three with One rather than a, which may be less confusing)

If we use base 2, or binary with a signed magnitude representation by using our first bit as an indication of "is this number negative" and the time we stop speaking to represent our most significant bit for example:
$-32 =$ a O O O O O a $=- (0\times 2^0 + 0\times 2^1 + 0\times 2^2 + 0\times 2^3 + 0\times 2^4 + 1\times 2^5)$

$-7 =$ a a a a
$=- (1\times 2^0 + 1\times 2^1 + 1\times 2^2)$

$-4 =$ a O O a
$=- (0\times 2^0 + 0\times 2^1 + 1\times 2^2)$

$0 =$ O O (or a O)
$=+ (0\times 2^0)$ (or $=- (0\times 2^0)$)

$17 =$ O a O O O a
$=+ (0\times 2^0 + 0\times 2^1 + 0\times 2^2 + 0\times 2^3 + 1\times 2^4)$

Note that

$2^n$ just means $2$ multiplied by itself $n$ times, so $2^3$ is $2\times 2 \times 2=8$
...and that the identity element of the multiplication group is $1$ so $2^0$ is $2$ multiplied by itself $0$ times, which must be $1$.

ONE ON ONE = 2

ONE OFF ONE = 0

ONEF: total 4 letters

by replacing plus and minus with on and off which mean the same.

Five letters: {Z, E, R, O, N}

No maths required, but you have to understand binary. We spell out the bits involved in a binary representation of the desired integer. The first bit is the sign bit (ZERO for positive, ONE for negative).

Examples:

-3: ONE ONE ONE
-2: ONE ONE ZERO
-1: ONE ONE
 0: ZERO ZERO
 1: ZERO ONE 
 2: ZERO ONE ZERO
 3: ZERO ONE ONE