There's a unfortunate typo in Silverman's formula for $b_2$: it should be $b_2 = a_1^2 + 4 a_2$. See here for a list of errata. Using the corrected formula, I get $c_4 = -80 = -2^4 \cdot 5$, which confirms that the curve has additive reduction at $p = 5$.
But there's no need to trust formulas in a book: we can simply reduce the equation for the curve modulo $p$ and $q$ and check if the singularities are nodes or cusps, as dalbouvet has begun to do in the comments. To do this, we first find the singular points. Letting
$$
F = y^2+y-(x^3-x^2+2x-2) \, ,
$$
these are exactly the solutions to $F = F_x = F_y=0$ where $F_x$ and $F_y$ are the specified partial derivatives. Let $(a,b)$ be the singular point. Making the change of variable $x \mapsto x - a, y \mapsto y - b$ moves the singularity to the origin, and then we can simply read off the tangent lines.
For $p = 5$, I find the singular point $(x,y) = (2,-3) = (2,2)$ and centered Weierstrass equation $y^2 = x^3$, which has a cusp. For $q = 7$, I find the singular point $(x,y) = (-3,-4) = (4,3)$ and centered Weierstrass equation $y^2 = x^3 - 3x^2$, which has a node.
Here are SageMathCells that compute do the above for $p = 5$ and $q = 7$ using the following code.
R.<x,y> = PolynomialRing(GF(5),2,order="lex")
f = x^3 - x^2 + 2*x - 2
F = y^2 + y - f
Fx = F.derivative(x)
Fy = F.derivative(y)
I = ideal([F,Fx,Fy])
show(I)
show(I.groebner_basis())
show(F(x=x+2,y=y-3))
${}$
R.<x,y> = PolynomialRing(GF(7),2,order="lex")
f = x^3 - x^2 + 2*x - 2
F = y^2 + y - f
Fx = F.derivative(x)
Fy = F.derivative(y)
I = ideal([F,Fx,Fy])
show(I)
show(I.groebner_basis())
show(F(x=x-3,y=y-4))
As a note, to compute the conductor of your elliptic curve you may find Tate's algorithm useful.
So like you said you want to find $p,c_4,c_6$ such that (assuming $p\ne 2,3$)
- $c_4\ne0$ and $c_6 \ne 0$
- $c_4^3 -c_6^2 \ne 0$
- $p|(c_4^3 -c_6^2)$
- $3v_p(c_4)=v_p(c_4^3)\ge v_p(c_4^3 -c_6^2)$ (integrality of $j$)
One way to get 3. to hold is for both $c_4,c_6$ to be divisible by $p$.
If $3v_p(c_4) =v_p(c_4^3) > v_p(c_6^2)=2v_p(c_6)$ then $v_p(c_4^3 -c_6^2) = v_p(c_6^2)$ so we get 4. for free.
So how to get $3v_p(c_4) > 2v_p(c_6)$ but $v_p(c_4)\ge 1$ and $v_p(c_6)\ge 1$, we can just take both to have valuation 1!
So why not try $c_4 = c_6 = p$, this satisfies all properties!
So for example $E : y^2 = x^3 -27\cdot5 x -54\cdot 5$ which is https://www.lmfdb.org/EllipticCurve/Q/10800dg1/
Best Answer
If the coefficient $a_1$ is even, then $$b_2=a_1^2+4a_2\equiv0\pmod{4} \qquad\text{ and }\qquad b_4=2a_4+a_1a_3\equiv0\pmod{2}.$$ It follows that $$\Delta=-b_2^2b_8-8b_4^3-27b_6^2+9b_2b_4b_6\equiv5b_6^2\pmod{8}.$$ This can never be congruent to $\pm1\pmod{8}$. Hence $a_1$ is odd, from which point you say you can conclude the proof.
Alternatively, as noted in the comments any elliptic curve over $\Bbb{Q}$ has a Weierstrass form $$y^2=x^3+ax+b,$$ with discriminant $\Delta=-16(4a^3+27b^2)$, which cannot equal $\pm1$ as it is a multiple of $16$.