You cannot do it in a single pass. Consider the set of all graphs of the following form: the vertices are ${s,t} cup A cup B$, where $|A|=|B|=n/2-1$. The only allowed edges are between $s$ and $A$, between $A$ and $B$, and between $B$ and $t$.

Suppose that the algorithm is first presented with the $(A,B)$ edges, and then with the $(s,A),(B,t)$ edges. After reading the $(A,B)$ edges, it must “remember” all of them, since there could be only one $(s,A)$ edge and only one $(B,t)$ edge, in which case the answer depends on whether the corresponding vertices in $A$ and $B$ are connected. Since there are $2^{(n/2-1)^2}$ choices for the $(A,B)$ edges, the algorithm must have a memory of at least $(n/2-1)^2$ bits (since after reading the $(A,B)$ edges, it could be in any of $2^{(n/2-1)^2}$ possible states).