Yes, nonsingularity can be defined purely algebraically, and you are correct that the corresponding algebraic notion is normality a.k.a. integral closedness. The equivalence, mentioned by user26857 in the comments, of this algebraic criterion for regularity with the more elementary Jacobian criterion is proved in Hartshorne's Algebraic Geometry, Theorem 5.1, p. 32.
Theorem 5.1. Let $Y \subseteq \mathbb{A}^n$ be an affine variety. Let $P \in Y$ be a point. Then $Y$ is nonsingular at $P$ if and only if the local ring $\mathcal{O}_{Y,P}$ is a regular local ring.
There are a slew of equivalent conditions for being a regular local ring, which are stated in Proposition 9.2 of Atiyah and MacDonald. Here's an abbreviated version of that result.
Proposition. Let $A$ be a Noetherian local domain of dimension one, $\mathfrak{m}$ its maximal ideal, $k = A/\mathfrak{m}$ its residue field. Then the following are equivalent:
(i) $A$ is a discrete valuation ring;
(ii) $A$ is normal, i.e., integrally closed;
(iii) $\mathfrak{m}$ is a principal ideal;
(iv) $\operatorname{dim}_k(\mathfrak{m}/\mathfrak{m}^2) = 1.$
Condition (iv) has a nice geometric interpretation. The $k$-vector space $\mathfrak{m}/\mathfrak{m}^2$ is the Zariski cotangent space. Since we are dealing with a curve, we would expect the tangent space (and hence the cotangent space) to have dimension 1, corresponding to the unique tangent line at a smooth point. However, at a singular point, such as the point $(0,0)$ in your example, the cotangent space is "too big": it has dimension $2$, corresponding to the two tangent lines at the origin.
You're definitely correct in thinking that the reason that $A_\mathfrak{m}$ is not integrally closed for $\mathfrak{m} = (x,y)$ is because the curve $C: y^2 = x^3 + x^2$ is singular at the point $(0,0)$. You could solve your exercise by appealing to geometric results, but it requires the above machinery, and I think it would be more instructive to try to solve it algebraically.
One note: $A_\mathfrak{m}$ can't be isomorphic to $\mathbb{C}[t]$ because $\mathbb{C}[t]$ is not local. Otherwise, you seem to be on the right track. You may also find the parametrization $x = t^2-1, y = t(t^2-1)$ helpful.
EDIT:
Here's more on your specific problem. The parametrization above gives an injective map
$$
A=\frac{\mathbb{C}[x,y]}{(y^2 - x^3 - x^2)} \to \mathbb{C}[t^2 - 1, t(t^2-1)] \subseteq \mathbb{C}[t] \, .
$$
If we localize $A$ at the ideal $(x - a, y - b)$, then this map has inverse $t \mapsto y/x$ for $a \neq 0$, so I thought the we could reduce the problem to localizing a one-variable polynomial ring. However, I had trouble working out the details.
Instead, let's show that $A_\mathfrak{m}$ is a regular local ring for every maximal ideal $\mathfrak{m} = (x-a, y-b)$ with $a \neq 0$ by using condition (iii) of the above proposition.
We can take any line that intersects the curve transversely (i.e., is not tangent) at $(a,b)$ as a generator for $\mathfrak{m}$. To show this, let's first consider the point $(-1,0)$ with maximal ideal $(x+1,y)$. Since $x$ is a unit in the local ring $A_{(x+1,y)}$ (it does not vanish at $(-1,0)$), then we have
$$
y^2 = x^2(x+1) \implies x+1 = \frac{y^2}{x^2} \in (y)
$$
so $(x+1,y) = (y)$ and the ideal is principal. We could have also seen this by considering the Taylor expansion of the function $f = x^3 + x^2 - y^2$ about $(-1,0)$:
$$
0 = f = (x+1) - 2(x+1)^2 - y^2 + (x+1)^3
$$
and rearranging, we get
$$
y^2 = (x+1) - 2(x+1)^2 + (x+1)^3 = (x+1)\overbrace{[1 - 2(x+1) + (x+1)^2]}^\text{unit}
$$
which again shows that $x+1 \in (y)$. The key here was that the partial derivative $f_x(-1,0) = 1 \neq 0$.
Since the only point of the curve with a vertical tangent line is $(-1,0)$, we can take $x-a$ as a generator for every $a \neq -1, 0$. This means that $f_y(a,b) \neq 0$ for all such $a$, so the Taylor expansion for $f$ around $(a,b)$ looks like
$$
0 = f = f_x(a,b) (x-a) + \overbrace{f_y(a,b)}^{\neq 0} (y-b) + \cdots \, .
$$
Using this expansion, we can use the same method as above:
(1) Group all the terms with a factor of $(y-b)$ together and pull a factor of $(y-b)$ out. Since $f_y(a,b) \neq 0$, the other factor will be a unit in the local ring.
(2) Move all the terms that only involve $(x-a)$ to the other side. The equality then shows that $y-b \in (x-a)$, so $(x-a,y-b) = (x-a)$ is principal.
We can actually be very explicit if we want and compute the various partial derivatives, which shows that
$$
f(x,y) = (3a^2 + 2a) (x-a) - (2b) (y-b) + (3a+1)(x-a)^2 - (y-b)^2 + (x-a)^3
$$
is the Taylor expansion about $(a,b)$. This way you can see the above manipulations explicitly if you'd like.
