I've established by brute-force that it is not possible to get a series with $\{-1,0,1\}$ coefficients this way. The maximum one can get is having such coefficients for degrees up to 121. Here is one example that achieves this many $\{-1,0,1\}$ coefficients:

Numerator:
     
    1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 - x^8 - x^9 - x^10 - x^11 - x^12 - x^13 - x^14 - x^15 - x^16 - x^17 - x^18 - x^19 - x^20 - x^21 - x^22 - x^23 - x^24 - x^25 - x^26 - x^27 - x^28 - x^29 - x^30 - x^31 + x^32 + x^33 + x^34 + x^35 + x^36 + x^37 + x^38 + x^39 + x^40 + x^41 + x^42 + x^43 + x^44 + x^45 + x^46 + x^47 + x^48 + x^49 + x^50 + x^51 + x^52 + x^53 + x^54 + x^55 + x^56 + x^57 + x^58 - x^59 + x^60 - x^61 + x^62 - x^63 - x^64 - x^65 - x^66 + x^67 + x^68 - x^69 + x^70 + x^71 + x^72 + x^73 + x^74 + x^75 + x^76 + x^77 + x^78 - x^79 - x^80 - x^81 - x^82 - x^83 - x^84 - x^85 + x^86 + x^87 - x^88 + x^89 + x^90 - x^91 + x^92 - x^93 - x^94 - x^95 + x^96 - x^97 - x^98 - x^99 + x^100 + x^101 - x^102 + x^103 + x^104 + x^105 - x^106 - x^107 + x^108 - x^109 + x^110 + x^111 - x^112 - x^113 + x^114 - x^115 + x^116 - x^117 - x^118 + x^119 - x^120 + x^121 + O(x^122)

Denominator:

    (1+x^2)*(1+x^3)*(1+x^4)*(1+x^5)*(1+x^6)*(1-x^7)*(1+x^8)*(1-x^9)*(1+x^10)*(1+x^11)*(1+x^12)*(1+x^13)*(1+x^14)*(1-x^15)*(1+x^16)*(1-x^17)*(1+x^18)*(1+x^19)*(1+x^20)*(1+x^21)*(1+x^22)*(1-x^23)*(1+x^24)*(1-x^25)*(1+x^26)*(1+x^27)*(1+x^28)*(1+x^29)*(1+x^30)*(1-x^31)*(1+x^32)*(1-x^33)*(1+x^34)*(1+x^35)*(1+x^36)*(1+x^37)*(1+x^38)*(1-x^39)*(1+x^40)*(1-x^41)*(1+x^42)*(1+x^43)*(1+x^44)*(1+x^45)*(1+x^46)*(1-x^47)*(1+x^48)*(1-x^49)*(1+x^50)*(1+x^51)*(1+x^52)*(1+x^53)*(1+x^54)*(1-x^55)*(1+x^56)*(1-x^57)*(1+x^58)*(1-x^59)*(1+x^60)*(1+x^61)*(1+x^62)*(1-x^63)*(1+x^64)*(1-x^65)*(1+x^66)*(1+x^67)*(1+x^68)*(1+x^69)*(1+x^70)*(1-x^71)*(1+x^72)*(1-x^73)*(1+x^74)*(1+x^75)*(1+x^76)*(1+x^77)*(1+x^78)*(1-x^79)*(1+x^80)*(1-x^81)*(1+x^82)*(1-x^83)*(1+x^84)*(1+x^85)*(1+x^86)*(1-x^87)*(1-x^88)*(1+x^89)*(1+x^90)*(1-x^91)*(1+x^92)*(1+x^93)*(1+x^94)*(1-x^95)*(1-x^96)*(1+x^97)*(1+x^98)*(1-x^99)*(1+x^100)*(1+x^101)*(1-x^102)*(1+x^103)*(1-x^104)*(1+x^105)*(1-x^106)*(1-x^107)*(1+x^108)*(1+x^109)*(1-x^110)*(1-x^111)*(1-x^112)*(1+x^113)*(1-x^114)*(1+x^115)*(1-x^116)*(1+x^117)*(1-x^118)*(1-x^119)*(1+x^120)*(1-x^121) + O(x^122)

Resulting series:

    1 + x - x^3 - x^4 - x^5 - x^6 + x^7 - x^12 - x^13 + x^14 + x^16 + x^17 + x^18 - x^20 + x^23 + x^24 - x^29 + x^32 + x^33 - x^35 - x^36 - x^37 - x^38 + x^39 + x^41 - x^43 - x^44 + x^48 + x^49 - x^51 - x^52 - x^53 - x^54 + x^56 + x^60 - x^61 - x^63 + x^67 - x^68 + x^69 + x^71 + x^72 + x^73 - x^76 + x^77 + x^81 + x^82 + x^83 - x^85 + x^87 - x^88 - x^89 + x^90 - x^91 + x^92 - x^93 - x^95 - x^99 - x^102 - x^104 + x^105 + x^107 - x^111 - x^114 + x^115 + x^118 - x^119 + x^121 + O(x^122)

**P.S.** It can be seen that in any series obtained this way, the coefficient of $x^n$ is congruent to $p(n)$ modulo 2. Hence, when the coefficient of $x^n$ belongs to $\{-1,0,1\}$, it is zero if and only if $n$ belongs to https://oeis.org/A001560

----

Here is my PARI/GP code for the brute-force:

    { test2(n,p,m) = my(q,t); if(n>=r, r=n;print(r-1," ",vecextract(s,2^(r-1)-1)); ); for(j=0,1, q=p/(1+(-1)^j*x^n); for(i=0,1, if( abs( polcoeff((m+(-1)^i*x^n + O(x^(n+1)))*q,n) )<=1, s[n]=[(-1)^i,(-1)^j]; test2(n+1,q,m+(-1)^i*x^n)); )); }
    s=vector(130); r=0; test2(2,1+O(x^131),1+x)

It tries to extend solutions by adding large and large powers and prints out current records as vectors of pairs of signs (one for numerator, one for denominator) for each degree. The series precision bound of 130 is chosen knowing that the degree won't go above 122 (otherwise it would raise an exception in PARI/GP).