Denote the lie-algebra of $SU(2)$ by $\mathfrak{su}(2)$, and that of $SO(3)$ by $\mathfrak{so}(3)$. Is the fact that both $\mathfrak{su}(2)$ and $\mathfrak{so}(3)$ have the same structure constants enough to ensure that they are isomorphic? I.e if $T_{\alpha},T_{\beta}$ are two generators in either: $$ [T_{\alpha},T_{\beta}] = \epsilon_{\alpha\beta}^{\gamma}T_{\gamma}$$
I can see that there's an argument that they are homomorphic.If they both have the same adjoint representation (same structure constants) then there is an isomorphism between the adjoint representation of $\mathfrak{su}(2)$ and $\mathfrak{so}(3)$. The adjoint representation of any lie algebra is a homomorphism (but not an isomorphism?) on that lie algebra.
So, since the composition of a homomorphism is itself a homomorphism (and being a homomorphism is a necessary condition to being an isomorphism) - if the structure constants are equal, then $\mathfrak{su}(2)$ and $\mathfrak{so}(3)$ are homomorphic.
Is there an extra ingredient I need to prove the isomoprhism - and am I roughly along the right lines?
Let $\mathfrak g$ and $\mathfrak h$ be two Lie algebras with the same dimension $n\in\mathbb N$. Let $\{e_1,\ldots,e_n\}$ and $\{f_1,\ldots,f_n\}$ be basis of $\mathfrak g$ and $\mathfrak h$ respectively. Suppose that there are numbers $\epsilon_{ij}^k$ such that$$(\forall i,j\in\{1,2,\ldots,n\}):[e_i,e_j]=\sum_{k=1}^n\epsilon_{ij}^ke_k\text{ and }[f_i,f_j]=\sum_{k=1}^n\epsilon_{ij}^kf_k.$$I suppose that this is what you mean when you assert that $\mathfrak g$ and $\mathfrak h$ “have the same structure constants”. If so, then, yes, $\mathfrak g$ and $\mathfrak h$ are isomorphic. You just take the (only) linear map $\psi\colon\mathfrak{g}\longrightarrow\mathfrak h$ such that$$(\forall i\in\{1,2,\ldots,n\}):\psi(e_i)=f_i.$$Then $\psi$ is an isomorphism.