Is there a way to both check a macro is defined and it equals a certain value at the same time

This may not work for the general case (I don't think there's a general solution to what you're asking for), but for your specific example you might consider changing this sequence of code:

#if(DEBUG_PRINT == 1)
    printf("%s", "Testing");
#endif

to:

if (DEBUG_PRINT == 1) {
    printf("%s", "Testing");
}

It's no more verbose and will fail to compile if DEBUG_PRINT is not defined or if it's defined to be something that cannot be compared with 1.

as far as I can tell, there is no preprocessor option to throw errors/warnings if a macro is not defined when used inside a #if statement.

It can't be an error because the C standard specifies that behavior is legal. From section 6.10.1/3 of ISO C99 standard:

After all replacements due to macro expansion and the defined unary operator have been performed, all remaining identifiers are replaced with the pp-number 0....

As Jim Balter notes in the comment below, though, some compilers (such as gcc) can issue warnings about it. However, since the behavior of substituting 0 for unrecognized preprocessor tokens is legal (and in many cases desirable), I'd expect that enabling such warnings in practice would generate a significant amount of noise.

There's no way to do exactly what you want. If you want to generate a compilation failure if the macro is not defined, you'll have to do it explicitly

#if !defined DEBUG_PRINT
#error DEBUG_PRINT is not defined.
#endif

for each source file that cares. Alternatively, you could convert your macro to a function-like macro and avoid using #if. For example, you could define a DEBUG_PRINT macro that expands to a printf call for debug builds but expands to nothing for non-debug builds. Any file that neglects to include the header defining the macro then would fail to compile.

---

Edit:

Regarding desirability, I have seen numerous times where code uses:

#if ENABLE_SOME_CODE
...
#endif

instead of:

#ifdef ENABLE_SOME_CODE
...
#endif

so that #define ENABLE_SOME_CODE 0 disables the code rather than enables it.

Rather than using DEBUG_PRINT directly in your source files, put this in the header file:

#if !defined(DEBUG_PRINT)
    #error DEBUG_PRINT is not defined
#endif

#if DEBUG_PRINT
    #define PrintDebug([args]) [definition]
#else
    #define PrintDebug
#endif

Any source file that uses PrintDebug but doesn't include the header file will fail to compile.

If you need other code than calls to PrintDebug to be compiled based on DEBUG_PRINT, consider using Michael Burr's suggestion of using plain if rather than #if (yes, the optimizer will not generate code within a false constant test).

Edit: And you can generalize PrintDebug above to include or exclude arbitrary code as long as you don't have commas that look like macro arguments:

#if !defined(IF_DEBUG)
    #error IF_DEBUG is not defined
#endif

#if IF_DEBUG
    #define IfDebug(code) code
#else
    #define IfDebug(code)
#endif

Then you can write stuff like

IfDebug(int count1;)  // IfDebug(int count1, count2;) won't work
IfDebug(int count2;)
...
IfDebug(count1++; count2++;)