Let $d > 0$ be a square-free integer.
If $d \equiv 2,3 \text{ mod } 4$, then $\mathbb{Z}[\sqrt{d}]$ is the ring of integers of the number field $\mathbb{Q}(\sqrt{d})$. We can thus consider the two maps $i_1, i_2 \colon \mathbb{Z}[\sqrt{d}] \to \mathbb{R}$ given by $x + y\sqrt{2} \mapsto x \pm y \sqrt{2}$ and the image of the induced map $$ SL_2(\mathbb{Z}[\sqrt{d}]) \hookrightarrow SL_2(\mathbb{R}) \times SL_2( \mathbb{R}) $$ is a lattice.
Does this also hold for the case $d \equiv 1 \text{ mod } 4$, where $\mathbb{Z}[\sqrt{d}]$ is in general not a ring of integers of a number field anymore?
Sure; if $A$ is the ring of integers of a number field and $B$ is an order then $\mathrm{SL}_2(B)$ has finite index in the version for $A$, since it contains the kernel of the map to the finite group $\mathrm{SL}_2(A/N)$ for some $N$ since $B$ will contain $NA$ where $N$ is the index. Of course you need to know that $\mathrm{SL}_2(A)$ is a lattice in $\mathrm{SL}_2(A \otimes \mathbf{R})$ which you implicitly seem to know but is also in standard books on arithmetic groups like the one by Dave Morris.