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.
Write
$$\pi = \frac{\mathrm{d}x}{2y+a_1x+a_3}=\frac{\mathrm{d}y}{3x^2+2a_2x+a_4-a_1y}\text{.}$$
Then $$\pi'=u\pi\text{,}$$
so that the expressions that are actually invariant are certain so-called "modular differential forms", viz., modular forms like
$$\begin{align}
\gamma_4 &= c_4 \pi^{\otimes 4} \\
\gamma_6 &= c_6 \pi^{\otimes 6} \\
\gamma_{12} &= \Delta \pi^{\otimes 12}
\end{align}$$
and quasi-modular forms like
$$\begin{align}
\gamma_2 &=(12x+b_2)\pi^{\otimes 2} \\
\gamma_3 &=(2y+a_1x+a_3)\pi^{\otimes 3}\text{.}
\end{align}$$
And while it's true that the coefficients of these expressions alone are not genuinely invariant under the law given for the Weierstrass form of curves, they are still "invariant up to powers": if, for two Weierstrass forms $E$, $E'$ over $\mathbb{Q}$, we calculate $c_4$ and $c'_4$, and we find that $c_4/c'_4$ is not of the form $u^4$ for any $\mathbb{Q}$, then we can conclude that $E$ and $E'$ are not isomorphic over $\mathbb{Q}$. To put it another way, we can imagine the "true invariants" of curves over $k$ to be orbits under certain actions of the multiplicative group $k^{\times}$:
$$\begin{align}[a_1]&\in (k\mod 2)/k^{\times}\\
[b_2]&\in(k\mod 12)/(k^{\times})^3 \\
[c_4]&\in k/(k^{\times})^4 \\
[c_6]&\in k/(k^{\times})^6 \\
[\Delta]&\in k/(k^{\times})^{12} \\
\end{align}$$
Best Answer
I put the sequence of Weierstrass coefficients $[a_1,a_2,a_3,a_4,a_6]=[0,1,-3,0,0]$ into the LMFDB search engine, and it gave the link to the curve $$ y^2+y=x^3+x^2+2.\qquad (*) $$ Apparently this is the minimal Weierstrass model. If you substitute $y\mapsto y-2$ into $(*)$ you get your equation back. Anyway inverting that subsitution gives you the list of integer points on your curve $$ (-1,3), (0,3), (9,30) $$ and the negatives of those points.