Lattice and isomorphim class of elliptic curve corresponds bijectively?
If we give lattice, we can define weierstrass form of elliptic curve, and gain one isomorphism class of elliptic curve.
Does converse holds? That is, if we give isomorphism class of elliptic curve, the lattice is unique?
There is a bijection between the set elliptic curves over $\mathbb{C}$, up to isomorphism, and the set of (full) lattices in $\newcommand{\C}{\mathbb{C}} \C$, up to homomthety: i.e., a lattice $\Lambda$ is equivalent to $c \Lambda$ for all $c \in \C^\times$. See Proposition I.4.4 of Silverman's Advanced Topics in the Arithmetic of Elliptic Curves for a proof.
To see that this homothety equivalence is necessary, consider how we get a lattice from an elliptic curve $E$. Since $E$ has genus $1$, then the set $\Omega(E)$ of holomorphic differentials on $E$ is a $1$-dimensional $\C$-vector space; let $\omega$ be a basis. Since $E(\C)$ is a complex torus, then the first homology group $\newcommand{\Z}{\mathbb{Z}} H^1(E(\C), \mathbb{Z})$ is a free abelian group of rank $2$; let $\gamma_1, \gamma_2$ be a basis.
Letting $$ \omega_1 := \int_{\gamma_1} \omega \qquad \text{and} \qquad \omega_2 := \int_{\gamma_2} \omega \, , $$ then $\Lambda := \Z \omega_1 \oplus \Z \omega_2$ is a lattice. We made two choices is defining $\Lambda$: the choice of $\omega$, and the choice of basis $\gamma_1, \gamma_2$. Picking a different basis $\gamma_1', \gamma_2'$ for $H^1(E(\C), \Z)$ will produce different periods $\omega_1'$ and $\omega_2'$, but they will still generate the same lattice, so this choice doesn't change $\Lambda$.
However, choosing a different differential $\omega'$ will change the lattice. Since $\Omega(E)$ is $1$-dimensional, then $\omega' = c \omega$ for some $c \in \C^\times$. Thus integrating $\omega' = c \omega$ to compute the periods yields $c \omega_1$ and $c \omega_2$, so we obtain the scaled lattice $c \Lambda$.
Allowing rescalings can be quite convenient as it allows us to normalize the period lattice. Given a lattice $\Lambda = \Z \omega_1 \oplus \Z \omega_2$, then $(1/\omega_1) \Lambda := \Z \oplus \Z (\omega_2/\omega_1)$ is equivalent. Thus we can always find a homothetic lattice one of whose generators is $1$. By swapping $\omega_1$ and $\omega_2$ if necessary, we can also assume that $\operatorname{Im}\omega_2/\omega_1 > 0$, so $\omega_2/\omega_1$ is in the upper half-plane. Thus for any elliptic curve $E$, we can find a period lattice $\Lambda$ of the form $\Lambda = \Z \oplus \Z \tau$ with $\tau \in \mathfrak{H}$.