Yes, it can be ignored. This is the casus irreducilbilis, in which the solutions to a cubic can be expressed only in terms of n-th roots that have complex values. The small imaginary parts are the result of round-off error.

If you'd like to get rid of them and don't mind Root[] objects, specify Cubics -> False on the exact system:

Solve[
   Rationalize[Rationalize[
     0 == 
      10125/(14818 \[Phi]) + 
       150 y0 (-6308.297007407409` - 148180.` \[Phi]) + 
       150 (946244.5511111114` + 13000.850814965874`/\[Phi] + 
          1.11135`*^7 \[Phi] - (
          0.022395039313571143` (6959.` + (337500.` - 
                2250.` y0) \[Phi])^(3/2))/\[Phi])
     ], 0]
   , {y0}, Cubics -> False
   ] /. Rationalize[\[Phi] -> 0.00145] // N

Solve::nongen: There may be values of the parameters for which some or all solutions are not valid.

(* {{y0 -> -13.9237}, {y0 -> 13.8775}, {y0 -> 1712.28}}  *)

Note that in this case Solve[] warns us that not all solutions may be valid, which indeed turned out to be the case. Root[] computes the real roots of a polynomial as Real numbers without complex round-off errors.

If you want a more rigorous method, use Method -> Reduce and increase MaxExtraConditions. Here MaxExtraConditions -> 1 is enough, but I started with MaxExtraConditions -> Infinity:

solRat = Solve[
   Rationalize[Rationalize[
     0 == 
      10125/(14818 \[Phi]) + 
       150 y0 (-6308.297007407409` - 148180.` \[Phi]) + 
       150 (946244.5511111114` + 13000.850814965874`/\[Phi] + 
          1.11135`*^7 \[Phi] - (
          0.022395039313571143` (6959.` + (337500.` - 
                2250.` y0) \[Phi])^(3/2))/\[Phi])
     ], 0]
   , {y0}, Cubics -> False, Method -> Reduce, MaxExtraConditions -> 1
   ]; (* 7 solutions depending on cases for the parameter phi *)
solPhi =(* select the defined cases *)
  Select[% /. Rationalize[\[Phi] -> 0.00145] // RootReduce, 
   NumericQ[y0 /. #] &];
solPhi // N (* we see the original was correct *)

Solve::useq: The answer found by Solve contains equational condition(s) {0==(192677381803147327908319230731396042083840204210666350+<<22>>)/(246032254379437568608800 (9387804582563199208933776766+<<1>>+138587276555703258030203175 Root[<<1>>&,1])),0==<<1>>/(2<<21>>00<<1>><<1>>),0==<<1>>/(246032254379437568608800<<1>><<1>>)}. A likely reason for this is that the solution set depends on branch cuts of Wolfram Language functions.

(*  {{y0 -> -13.9237}, {y0 -> 13.8775}}  *)

This time we get more of an idea why the previous Solve[] said not all solutions may be valid. We also see that several other solution-cases were rejected by the OP's original Solve[].

In the OP's second code, increasing MaxExtraConditions yields conditions on the solutions that are helpful in filtering out bogus solutions:

solApprox = 
  Solve[0 == 
    10125/(14818 \[Phi]) + 
     150 y0 (-6308.297007407409` - 148180.` \[Phi]) + 
     150 (946244.5511111114` + 13000.850814965874`/\[Phi] + 
        1.11135`*^7 \[Phi] - (
        0.022395039313571143` (6959.` + (337500.` - 
              2250.` y0) \[Phi])^(3/2))/\[Phi]), y0,
   Cubics -> False, MaxExtraConditions -> 1];
solApprox /. \[Phi] -> 0.00145
(* <warnings omitted>
{{y0 -> Undefined}, {y0 -> Undefined}, {y0 -> Undefined}, {y0 -> Undefined},
 {y0 -> -13.9237}, {y0 -> 13.8775}, {y0 -> Undefined}}
*)

If we remove the conditions from the solutions with Normal[], we see the last three are the solutions returned by the OP's second Solve[]:

Normal@solApprox /. \[Phi] -> 0.00145
(*
{{y0 -> -68.88}, {y0 -> -67.7773}, {y0 -> 41.1304 - 66.9167 I}, {y0 -> 41.1304 + 66.9167 I},
 {y0 -> -13.9237}, {y0 -> 13.8775}, {y0 ->1712.28}}
*)

But I wonder if even a small imaginary unit can be ignored?

The third solution is not valid for that value of the parameter:

eq0 = 0 == 
   10125/(14818 \[Phi]) + 
    150 y0 (-6308.297007407409` - 148180.` \[Phi]) + 
    150 (946244.5511111114` + 13000.850814965874`/\[Phi] + 
       1.11135`*^7 \[Phi] - (
       0.022395039313571143` (6959.` + (337500.` - 
             2250.` y0) \[Phi])^(3/2))/\[Phi]);
\[Phi]0 = 0.00145;
Solve[eq0, y0, Assumptions -> \[Phi] == \[Phi]0]
Solve[eq0 /. \[Phi] -> \[Phi]0, y0]
sol3 = Solve[eq0, y0] /. \[Phi] -> \[Phi]0
eq0[[2]] /. \[Phi] -> \[Phi]0 /. sol3

The formulas for the first and second solution contain an imaginary unit, but it is multiplied by a very small coefficient, probably negligible. The main problem is that all three solutions have fractional powers, which are problematic.