Consider a large 3d torus with even side lengths that contains the configuration of 5 starter dominoes. Colour the cells making up the torus white and black in an alternating 3d checkerboard pattern. (Each white cell has 6 black neighbours, and vice-versa. This is possible because the torus has even side lengths.)
Let $X$ be the set of black squares not covered by any of the 5 starter dominoes and $Y$ be the set of white squares (including the 5 covered by starter dominoes). Note that $|X|+5 = |Y|$. For every set $B\subset X$, define $N(B)$ to be the set of all neighbours of the elements of $B$. (So $y\in Y$ will be an element of $N(B)$ if and only if it is neighbours with some $x\in B$.)
According to Hall's marriage theorem, it will be possible to tile the torus with dominoes (and thus the whole plane, since the torus tiling can just be repeated) if the following condition is met: "For all sets $B\subset X$, $|N(B)| \geq |B| + 5$." i.e. each subset of $n$ black cells has at least $n+5$ white neighbours. (This is so that even if all 5 starter dominoes happen to be covering our neighbours, we still have $n$ neighbours left over to satisfy the Hall marriage criterion. This is a sufficient condition, but not a necessary one.)
This seems obviously true after a bit of thought, but I found it very difficult to come up with a proof. Here is a messy one with 3 separate cases. Maybe it's possible to come up with a better proof.
Make the torus very large: at least $4\cdot 21600$ cells long in each direction. (It may need to be bigger in order to contain the full configuration of starter dominoes which could be spaced very far apart, but that's fine for our proof.)
We will show that $|N(B)| \geq |B|+5$ for each of 3 cases.
Case 1: $|B| < 216000$
Write down the $x$ coordinates of all elements of $B$ and all neighbours of $B$ (i.e. elements of $N(B)$). If the $x$ coordinates of the elements of $B$ are $x_1, x_2, \dots$, then the coordinates of the elements and their neighbours are $x_1, x_1-1, x_1+1, x_2, x_2-2, x_2+2, \dots$. We've just written down less than $3|B| < 3\cdot 216000$ unique $x$ coordinates. Since the side length of the torus is at least $4\cdot 216000$, we can find an $x$ coordinate that belongs to no elements of $B$ and no neighbour of $B$. This defines of plane of cells that all have that as their $x$ coordinate and are neither contained in $B$ nor neighbours of $B$.
Using the same method, find unused $y$ and $z$ coordinates, which also define planes that do not contain any elements of $B$ or neighbours of $B$. Remove these 3 planes from the torus, leaving behind a large block, without the wraparound structure of the torus.
Now construct the set $B$ in layers, starting from a single cell that has no other elements of $B$ below it. This single cell has $6$ white neighbours. $6 \geq 5+1$, so this will serve as our base case. Add the elements of $B$ one at a time, always making sure to complete 1 layer before moving on to the next. Each black cell we add to $B$ always adds at least 1 new white cell to $N(B)$, namely the cell just above it. So we always have $|N(B)| \geq 5+|B|$ at each step of the process.
Case 2: $216000 \leq |B| \leq |Y|-216000$
Divide the space up into $2\times 1\times 1$ blocks. (Do this so that the edges of the blocks are all lined up: We map $(x,y,z)$ to $(x, y, \lfloor z/2 \rfloor)$, and any two cells mapped to the same final coordinates belong to the same block.) We say a block is full if its black cell is an element of $B$, and empty otherwise. An empty cell may or may not contain a white neighbour of $B$, but all full cells contain one. We just need to show that there are at least 5 empty cells containing white neighbours of $B$. Since there are between 216000 and $|Y|-216000$ full blocks, the boundary between full blocks and empty blocks must contain at least 60 faces. (Faces are just the sides of the blocks. A face can have a surface area of 1 or 2. If we shrink the blocks by a factor of 2 in the $z$ direction, so that the blocks become cubes, we're claiming that any set of between 216000 and $|Y|-216000$ unit cubes in a torus of size at least $216000\times216000\times108000$ must have a surface area of at least 60. This really needs a proof of its own, but it seems likely enough to be true, so I'll leave it as an "exercise for the reader".)
Since each empty block that neighbours a full block can have at most 6 of these faces as neighbours, there must be at least 10 empty blocks with full blocks as neighbours. Any empty block with a full block to the north, south east, or west must have its white cell be a neighbour of $B$ (its white cell touches the black cells of all horizontally adjacent blocks). As for vertical neighbours: In some columns of blocks, the white cell in a block is below the black cell. In that case all empty blocks with a full block underneath will contain a white neighbour of $B$. In other columns, the white cell is above the black cell in each block, so all empty blocks with a full block above will contain a white neighbour of $B$. Note that in each column, there will be exactly as many empty blocks with full blocks above them as there are empty blocks with full blocks below them. (Because runs of empty and full blocks must alternate and the torus loops around.) Therefore, if there are $n$ empty blocks with full blocks as neighbours, at least $n/2$ empty blocks must contain white cells neighbouring $B$.
We now know that there are at least $10/2=5$ empty blocks containing white neighbours of $B$, which completes the proof for this case.
Case 3: $|Y|-216000 < |B|$
By the same argument as case 1, we can split the torus along 3 planes so that none of the planes contain any black cells not in $B$ or their neighbours. Then we start with the set of all black cells, and remove elements from it 1 at a time until we are left with $B$. As before, we proceed in layers. After removing 5 black cells, we have $|Y|$ neighbours, and $|Y|-5$ black cells. (We still have |Y| neighbours at this point because we started with all white cells being neighbours to the set of all black cells, and to get rid of a white cell as a neighbour, we'd need to remove all 6 black cells surrounding it, not just the 5 we have actually removed. Note that $|B| \leq |X| = |Y|-5$, so we always have to remove at least 5 black cells to get to $B$. So this will serve as our base case.
We continue removing cells 1 layer at a time. Each black cell we remove can only remove 1 neighbour when we take it away, namely the white cell directly below it, since to eliminate a white cell as a neighbour, we certainly must eliminate the black cell just above it first. And because we are proceeding in layers, the cell above will be the last of a white cell's neighbours to be removed. So the desired property is preserved with each cell we remove.
Commentary: This proof looks like it can be generalized to prove your conjecture in higher dimensions too.